Question 15 Marks
Evaluate the following integrals:
$\int(\text{x}+1)\sqrt{2\text{x}^2+3}\text{dx}$
AnswerLet $\text{I}=\int(\text{x}+1)\sqrt{2\text{x}^2+3}\text{dx}$
Let $\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(2\text{x}^2+3)+\mu$
$=\lambda(4\text{x})+\mu$
Equating similar terms, we get,
$4\lambda=1\ \Rightarrow\ \lambda=\frac{1}{4}$
$\mu=1$
$\therefore\ \text{I}=\int\frac{1}{4}(4\text{x})\sqrt{2\text{x}^2+3}\text{dx}+\int1.\sqrt{2\text{x}^2+3}\text{dx}$
Let $2\text{x}^2+3=\text{t}$
$\Rightarrow4\text{x dx}=\text{dt}$
$\text{I}=\frac{1}{4}\int\sqrt{\text{t}}\text{dt}+\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}}\text{dx}$
$=\frac{1}{4}.\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\sqrt2\begin{Bmatrix}\frac{\text{x}}{2}\sqrt{\text{x}^2+\frac{3}{2}}\\+\frac{3}{4}\log\Big|\text{x}+\sqrt{\text{x}^2+\frac{3}{2}}\Big|+\text{C}\end{Bmatrix}$
Hence,
$\text{I}=\frac{1}{6}(2\text{x}^2+3)^{\frac{3}{2}}+\frac{\text{x}}{2}\sqrt{2\text{x}^2+3}\\+\frac{3}{2\sqrt2}\log\Big|\text{x}+\sqrt{\text{x}^2+\frac{3}{2}}\Big|+\text{C}$
View full question & answer→Question 25 Marks
$\int\frac{2-3\text{x}}{\sqrt{1+3\text{x}}}\text{dx}$
AnswerLet I $=\int\frac{2-3\text{x}}{\sqrt{1+3\text{x}}}\times\text{dx}.$ Then,
$\text{I}=\int\frac{2-3\text{x}-1+1}{\sqrt{1+3\text{x}}}\text{dx}$
$=\int\frac{-3\text{x}-1+3}{\sqrt{1+3\text{x}}}\text{dx}$
$=\int-\frac{(3\text{x}+1)}{\sqrt{1+3\text{x}}}\text{dx}+3\int\frac{1}{\sqrt{1+3\text{x}}}\text{dx}$
$=-1\int\frac{1+3\text{x}}{\sqrt{1+3\text{x}}}\text{dx}+3\int\frac{1}{\sqrt{1+3\text{x}}}\text{dx}$
$=-1\int(1+3\text{x})^\frac{1}{2}\text{dx}+3\int(1+3\text{x})^\frac{-1}{2}\text{dx}$
$=-1\times\frac{(1+3\text{x})^\frac{3}{2}}{\frac{3}{2}\times3}+3\times\frac{(1+3\text{x})^\frac{1}{2}}{\frac{1}{2}\times3}+\text{C}$
$=-\frac{2}{9}\times(1+3\text{x})^\frac{3}{2}+2(1+3\text{x})^\frac{1}{2}+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[-\frac{1}{9}(1+3\text{x})^1+1\Big]+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[\frac{-1-3\text{x}+9}{9}\Big]+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[\frac{8-3\text{x}}{9}\Big]+\text{C}$
$=\frac{2}{9}\sqrt{1+3\text{x}}(8-3\text{x})+\text{C}$
$\therefore\text{I}=\frac{2}{9}(8-9\text{x})\sqrt{1+3\text{x}}+\text{C}$
View full question & answer→Question 35 Marks
Evaluate the following integrals:
$\int \frac{\text{x}+1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
AnswerLet $\text{I}=\int \frac{\text{x}+1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
$\text{I}=\int\frac{(\text{x}-1)+2}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
$\text{I}=\int\frac{\text{dx}}{\sqrt{\text{x}+2}}+2\int\frac{\text{dx}}{(\text{x}+1)\sqrt{\text{x}+2}}\ ...(\text{i})$
Now, $\int\frac{\text{dx}}{\sqrt{\text{x}+2}}+2\sqrt{\text{x}+2}+\text{C}_1$
and, $\int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Let $\text{x}+2=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\therefore\ \int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}=2\int\frac{\text{t dt}}{(\text{t}^2-3)\text{t}}=2\int\frac{\text{dt}}{\text{t}^2-3}$
$=\frac{2\times1}{2\sqrt{3}}\log\bigg|\frac{\text{t}-\sqrt{3}}{\text{t}+\sqrt{3}}\bigg|+\text{C}_2$
$=\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}_2$
Thus, from (i)
$\text{I}=2\sqrt{\text{x}+2}+\text{C}_1+\frac{2}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}_2$
Hence, $\text{I}=2\sqrt{\text{x}+2}+\frac{2}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
View full question & answer→Question 45 Marks
$\int\frac{\text{x}^2}{\sqrt{1-\text{x}}}\text{dx}$
Answer$\int\frac{\text{x}^2}{\sqrt{1-\text{x}}}\text{dx}$
Let $1-\text{x}=\text{t}$
$\Rightarrow\text{x}=1-\text{t}$
$\Rightarrow1=-\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=-\text{dt}$
Now, $\int\frac{\text{x}^2}{\sqrt{1-\text{x}}}\text{dx}$
$=\int\frac{(1-\text{t})^2}{\sqrt{\text{t}}}\text{dt}$
$=\int\Big(\frac{1-\text{t}^2-2\text{t}}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\frac{1}{\sqrt{\text{t}}}+\frac{\text{t}^2}{\sqrt{\text{t}}}-\frac{2\text{t}}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\text{t}^{-\frac{1}{2}}+\text{t}^{\frac{3}{2}}-2\text{t}^{\frac{1}{2}}\Big)\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}-\frac{2\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\text{t}^{\frac{1}{2}}+\frac{2}{5}\text{t}^{\frac{5}{2}}-\frac{4}{3}\text{t}^{\frac{3}{2}}+\text{C}$
$=2\text{t}^{\frac{1}{2}}\Big[1+\frac{\text{t}^2}{5}-\frac{2}{3}\text{t}\Big]+\text{C}$
$=2\text{t}^{\frac{1}{2}}\Big[\frac{15+3\text{t}^2-10\text{t}}{15}\Big]+\text{C}$
$=2\sqrt{1-\text{x}}\Big[\frac{15+3(1-\text{x})^2-10(1-\text{x})}{15}\Big]+\text{C}$
$=\frac{2}{15}\sqrt{1-\text{x}}\big[15+3(1^2+\text{x}^2-2\text{x})-10+10\text{x}\big]+\text{C}$
$=\frac{2}{15}\sqrt{1-\text{x}}\big[15+3+3\text{x}^2-6\text{x}+10+10\text{x}\big]+\text{C}$
$=\frac{2}{15}\sqrt{1-\text{x}}\big[3\text{x}^2+4\text{x}+8\big]+\text{C}$
View full question & answer→Question 55 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}-1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
AnswerLet $\int\frac{\text{x}^2+\text{x}-1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}=\frac{\text{A}}{(\text{x}+1)}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{(\text{x}+2)}$$\Rightarrow\text{x}^2+\text{x}-1=\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}+1)^2$
$=(\text{A}+\text{C})\text{x}^2+(3\text{A}+\text{B}+2\text{C})\text{x}+(2\text{A}+2\text{B}+\text{C})$
Equating similar terms
$\text{A}+\text{C}=1,3\text{A}+\text{B}+2\text{C}=1,2\text{A}+2\text{B}+\text{C}=-1$
Solving, we get, A = 0, B = -1, C = 1
Thus,
$\text{I}=0\int\frac{\text{dx}}{\text{x}+1}+(-1)\int\frac{\text{dx}}{(\text{x}+1)^2}+1\int\frac{\text{dx}}{(\text{x}+2)}$
$=\frac{1}{\text{x}+1}+\log|\text{x}+2|+\text{C}$
$\text{I}=\frac{1}{\text{x}+1}+\log|\text{x}+2|+\text{C}$
View full question & answer→Question 65 Marks
Evaluate the following integrals:
$\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$We express $\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}(3-4\text{x}-\text{x}^2)+\text{B}$
$\text{x}+3=\text{A}(-4-2\text{x})+\text{B}$
Equating the co-efficient of x and constants, we get
$1=-2\text{A}$ and $3=-4\text{A + B}$
or $\text{A}=-\frac{1}{2}$ and $\text{B}=1$
$\therefore\ \text{I}=\int\Big(-\frac{1}{2}(-4-2\text{x})+1\Big)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
$=-\frac{1}{2}\int(-4-2\text{x})\sqrt{3-4\text{x}-\text{x}^2}\text{dx}+\int\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
$=-\frac{1}{2}\text{I}_1+\text{I}_2\ \dots(1)$
Now, $\text{I}_1=\int(-4-2\text{x})\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
Let $3-4\text{x}-\text{x}^2=\text{u}$
On differentiating both sides, we get
$(-4-2\text{x})\text{dx = du}$
$\therefore\ \text{I}_1=\int\sqrt{\text{u}}\text{du}$
$=\frac{2}{3}\text{u}^{\frac{3}{2}}+\text{c}_1$
$=\frac{2}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\text{c}_1\ \dots(2)$
And, $\text{I}_2=\int\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{3+4-4-4\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{(\sqrt7)^2-(\text{x}+2)^2}\text{dx}$
Let $(\text{x}+2)=\text{u}$
On differentiating both sides, we get
$\text{dx = du}$
$\therefore\ \text{I}_2=\int\sqrt{(\sqrt7)^2-(\text{u})^2}\text{du}$
$=\frac{\text{u}}{2}\sqrt{(\sqrt7)^2-(\text{u})^2}+\frac{1}{2}(\sqrt7)^2\sin^{-1}\Big(\frac{\text{u}}{\sqrt7}\Big)+\text{c}_2$
$=\frac{\text{x}+2}{2}\sqrt{7-(\text{x}+2)^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2$
$=\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2\ \dots(3)$
From (1), (2) and (3), we get
$\therefore\ \text{I}=-\frac{1}{2}\Big(\frac{2}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\text{c}_1\Big)\\+\Big(\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2\Big)$
$=-\frac{1}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}\\+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{C}$
Hence,
$\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}=-\frac{1}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}\\+\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{C}$
View full question & answer→Question 75 Marks
Evaluate the following intregals: $\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}+1)^2}\ \text{dx}$
AnswerLet $\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{(\text{x}-1)}+\frac{\text{C}}{\text{(x}+1)^2}$
$\Rightarrow x^2 = A(x + 1)^{2 }+ B(x - 1) (x + 1) + C(x - 1)$
$= (A + B) x^2 + (2A + C) x + (A - B - C)$
Equating similar terms,
$A + B = , 2A + C = 0, A - B - C = 0$
Solving, we get, $\text{A}=\frac{1}{4},\text{B}=\frac{3}{4},\text{C}=-\frac{1}{2}$
Thus,
$\text{I}=\frac{1}{4}\int\frac{\text{dx}}{\text{x}-1}+\frac{3}{4}\int\frac{\text{dx}}{\text{x}+1}-\frac{1}{2}\int\frac{\text{dx}}{(\text{x}+1)}^2$
$=\frac{1}{4}\log|\text{x}-1|+\frac{3}{4}\log|\text{x}+1|+\frac{1}{2(\text{x}+1)}+\text{C}$
$\text{I}=\frac{1}{4}\log|\text{x}-1|+\frac{3}{4}\log|\text{x}+1|+\frac{1}{2(\text{x}+1)}\text{C}$
View full question & answer→Question 85 Marks
Evaluate the following intregals:
$\int\frac{18}{(\text{x}+2)(\text{x}^2+4)}\text{ dx}$
AnswerLet $\frac{18}{(\text{x}+2)(\text{x}^2+4)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+4}$$\Rightarrow18=\text{A}(\text{x}^2+4)+(\text{Bx}+\text{C})(\text{x}+2)$
$18=(\text{A}+\text{B})\text{x}^2+(2\text{B}+\text{C})\text{x}+(4\text{A}+2\text{C})$
Equating similar terms, get,
$\text{A}+\text{B}=0,2\text{B}+\text{C}=0,4\text{A}+2\text{C}=18$
Solving, we get,
$\text{A}=\frac{9}{4},\text{B}=-\frac{9}{4},\text{C}=\frac{9}{2}$
Thus,
$\text{I}=\frac{9}{4}\int\frac{\text{dx}}{\text{x}+2}+\Big(-\frac{9}{4}\Big)\frac{\text{x}}{\text{x}^2+4}\ \text{dx}+\frac{9}{2}\int\frac{\text{dx}}{\text{x}^2+4}$
$\text{I}=\frac{9}{4}\log|\text{x}+2|-\frac{9}{8}\log|\text{x}^2+4|+\frac{9}{4}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$ $\Big[\because\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{9}\Big]$
View full question & answer→Question 95 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
AnswerWe have$\text{I}=\int\frac{(\text{x}^2+\text{x}+1)}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
Let, $\int\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{\text{x}+2}$
$\Rightarrow\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}+1)^2}{(\text{x}+1)^2(\text{x}+2)}$
$\Rightarrow\text{x}^2+\text{x}+1=\text{A}(\text{x}^2+\text{x}+2\text{x}+2)\\+\text{Bx}+2\text{B}+\text{C}(\text{x}^2+2\text{x}+1)$
$\Rightarrow\text{x}^2+\text{x}+1=(\text{A}+\text{C})\text{x}^2+(3\text{A}+\text{B}+2\text{C})\text{x}\\+(2\text{A}+2\text{B}+\text{C})$
Equating coefficient of like terms,
$\text{A}+\text{C}=1\ ...(1)$
$3\text{A}+\text{B}+2\text{C}=1\ ...(2)$
$2\text{A}+2\text{B}+\text{C}=1\ ...(3)$
Solving these three equation we get
$\text{A}=-2$
$\text{B}=1$
$\text{C}=3$
Hence, $\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{-2}{\text{x}+1}+\frac{1}{(\text{x}+1)^2}+\frac{3}{\text{x}+2}$
$\therefore\text{I}=-2\int\frac{\text{dx}}{\text{x}+1}+\int\frac{\text{d}}{(\text{x}+1)^2}+3\int\frac{\text{dx}}{\text{x}+2}$
$=-2\log|\text{x}+1|-\frac{1}{\text{x}+1}+3\log|\text{x}+2|+\text{C}$
View full question & answer→Question 105 Marks
Evaluate the following intregals:
$\int\frac{2\text{x}+1}{\sqrt{\text{x}^2+2\text{x}-1}}\ \text{dx}$
Answer Let $\text{I}\int\frac{2\text{x}+1}{\sqrt{\text{x}^2+2\text{x}-1}}\ \text{dx}$Let $2\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x}-1)+\mu$
$=\lambda(2\text{x}+2)+\mu$
$2\text{x}+1=(2\lambda)\text{x}+2\lambda+\mu$
Comapring the coefficient of like powers of x,
$2\lambda=2\Rightarrow\lambda=1$
$2\lambda+\mu=1\Rightarrow2(1)+\mu=1$
$\mu=-1$
So, $\text{I}=\int\frac{(2\text{x}+2)-1}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}$
$\text{I}=\int\frac{(2\text{x}+2)}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}-\int\frac{1}{\sqrt{\text{x}^2+2\text{x}+(1)^2-(1)^2-1}}$
$\text{I}=\int\frac{2\text{x}+2}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}-\int\frac{1}{\sqrt{(\text{x}+1)^2-(\sqrt{2}^2})}$
$\text{I}=(2\sqrt{\text{x}^2+2\text{x}-1})-\log\big|(\text{x}+1)+\sqrt{(\text{x}+1)^2-(\sqrt{2}})^2\big|+\text{C}$ $\Big[\text{since},\int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{x}^2-\text{a}^2}}\text{dx}=\log\big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\big|+\text{C}\Big]$
$\text{I}=2\sqrt{\text{x}^2+2\text{x}-1}-\log\big|\text{x}+1+\sqrt{\text{x}^2+2\text{x}-1}\big|+\text{c}$
View full question & answer→Question 115 Marks
Evaluate the following integrals: $\int\frac{\text{x}^2+9}{\text{x}^4+81}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+9}{\text{x}^4+81}\ \text{dx}$ Dividing numerator and denominator by $x^2$
$\text{I}=\int\frac{1+\frac{9}{\text{x}^2}}{\text{x}^2+\frac{81}{\text{x}^2}}\ \text{dx}$
$=\int\frac{1+\frac{9}{\text{x}^2}}{\Big(\text{x}-\frac{9}{\text{x}}\Big)^2+18}$
Let $\Big(\text{x}-\frac{9}{\text{x}}\Big)=\text{t}$
$\Rightarrow\Big(1+\frac{9}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+18}$
$\text{I}=\frac{1}{3\sqrt{2}}\tan^{-1}\Big(\frac{\text{t}}{3\sqrt{2}}\Big)+\text{C}$
Thus,
$\text{I}=\frac{1}{3\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}^2-9}{3\sqrt{2}}\Big)+\text{C}$
View full question & answer→Question 125 Marks
$\int\frac{\text{x}+1}{\sqrt{2\text{x}+3}}\text{dx}$
Answer$\int\Big(\frac{\text{x}+1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+2}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+3-1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+3}{\sqrt{2\text{x}+3}}-\frac{1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\sqrt{2\text{x}+3}-\frac{1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\Big[\int(2\text{x}+3)^\frac{1}{2}\text{dx}-\int(2\text{x}+3)^{-\frac{1}{2}}\text{dx}\Big]$
$=\frac{1}{2}\Bigg[\frac{(2\text{x}+3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\Big)}-\frac{(2\text{x}+3)^{-\frac{1}{2}+1}}{2\big(-\frac{1}{2}+1\big)}+\text{C}\Bigg]$
$=\frac{1}{2}\Big[\frac{1}{3}(2\text{x}+3)^\frac{3}{2}-(2\text{x}+3)^\frac{1}{2}+\text{C}\Big]$
$=\frac{1}{6}(2\text{x}+3)^\frac{3}{2}-\frac{1}{2}(2\text{x}+3)^\frac{1}{2}+\text{C}$
View full question & answer→Question 135 Marks
Evaluate the following integrals:$\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}$Let the first function be (\log x) and second function be $\frac{1}{(\text{x}+1)^2}.$
First we find the integral of the second function, i.e, $\int\frac{1}{(\text{x}+1)^2}\text{dx}.$
Put $t = (x + 1)^{ }$Then $dt = dx$
Therefore,
$\int\frac{1}{(\text{x}+1)^2}\text{dx}=\int\text{t}^{-2}\text{dt}$
$=-\frac{1}{\text{t}}$
$=-\frac{1}{1+\text{x}}$
Hence, using integration by parts, we get
$\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}=(\log\text{x})\int\frac{1}{(\text{x}+1)^2}\text{dx}-\int\Big[\Big(\frac{\text{d}(\log\text{x})}{\text{dx}}\Big)\int\frac{1}{(\text{x}+1)^2}\text{dx}\Big]\text{dx}$
$=(\log\text{x})\Big(-\frac{1}{1+\text{x}}\Big)-\int\big(\frac{1}{\text{x}}\big)\Big(-\frac{1}{1+\text{x}}\Big)\text{dx}$
$=-\frac{\log\text{x}}{1+\text{x}}+\int\Big(\frac{1}{\text{x}^2+\text{x}}\Big)\text{dx}$
$=-\frac{\log\text{x}}{1+\text{}x}+\int\frac{1}{\text{x}^2+\text{x}+\frac{1}{4}-\frac{1}{4}}\text{dx}$
$=-\frac{\log\text{x}}{1+\text{x}}+\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{dx}$
$=-\frac{\log\text{x}}{1+\text{x}}+\frac{1}{2\times\frac{1}{2}}\log\Bigg|\frac{\text{x}+\frac{1}{2}-\frac{1}{2}}{\text{x}+\frac{1}{2}+\frac{1}{2}}\Bigg|+\text{C}$
$=-\frac{\log\text{x}}{1+\text{x}}+\log\Big|\frac{\text{x}}{\text{x}+1}\Big|+\text{C}$
Hence, $\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}=-\frac{\log\text{x}}{1+\text{x}}+\log\Big|\frac{\text{x}}{\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 145 Marks
Evaluate the following integrals:
$\int\tan^{-1}(\sqrt{\text{x}})\text{dx}$
AnswerLet $\text{I}=\int\tan^{-1}(\sqrt{\text{x}})\text{dx}$Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\text{I}=\int2\text{t}\tan^{-1}\text{t dt}$
$=2\Big[\tan^{-1}\text{t}\int\text{t dt}-\int\Big(\frac{1}{1+\text{t}^2}\int\text{t dt}\Big)\text{dt}\Big]$
$=2\Big[\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\int\frac{\text{t}^2}{2(1+\text{t}^2)}\text{dt}\Big]$
$=\text{t}^2\tan^{-1}\text{t}-\int\frac{\text{t}^2+1-1}{1+\text{t}^2}\text{dt}$
$=\text{t}^2\tan^{-1}\text{t}-\int\Big(1-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$=\text{t}^2\tan^{-1}\text{t}-\text{t}+\tan^{-1}\text{t + C}$
$=(\text{t}^2+1)\tan^{-1}\text{t}-\text{t + C}$
$\text{I}=(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 155 Marks
Evaluate the following integrals: $\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
AnswerConsider the integral $\text{I}=\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
Let us express $\text{x}-3=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+3\text{x}-18)+\mu$
$\Rightarrow\text{x}-3=\lambda(2\text{x}+3)+\mu$
$\Rightarrow\text{x}-3=2\lambda\text{x}+3\lambda+\mu$
Comparing the co$-$efficients, we have,
$2\lambda=1\text{ and }3\lambda+\mu=-3$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }3\times\frac{1}{2}+\mu=-3$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }\mu-3-\frac{3}{2}$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }\mu=-\frac{9}{2}$
Then
$\text{x}-3=\lambda(2\text{x}+3)+\mu$
Now the integral $\text{I}=\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$=\int\Big(\frac{1}{2}(2\text{x}+3)-\frac{9}{2}\Big)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$\text{I}=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}\\-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$\Rightarrow\text{I}=\text{I}_1+\text{I}_2$
where, $\text{I}_1=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
and $\text{I}_2=-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
Let us consider the integral, $I_1:$
$\text{I}_1=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
Substituting, $\text{x}^2+3\text{x}-18=\text{t}$
$\Rightarrow(2\text{x}+3)\text{dx = dt}$
Thus,
$\text{I}_1=\frac{1}{2}\int\sqrt{\text{t}}\text{dt}$
$=\frac{1}{2}\times\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{1}{2}\times\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=\frac{1}{2}\times\frac{2}{3}\times\text{t}^{\frac{3}{2}}+\text{C}$
$=\frac{1}{3}\times\text{t}^{\frac{3}{2}}+\text{C}$
$=\frac{1}{3}\times(\text{x}^2+3\text{x}-18)^{\frac{3}{2}}+\text{C}$
Now consider the integral
$\text{I}_2=-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\text{x}^2+2\times\frac{3}{2}\text{x}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\frac{9}{4}-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{4}+18\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9+72}{4}\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{81}{4}\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\text{dx}$
We know that,
$\int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2-\text{a}^2}-\frac{1}{2}\text{a}^2\log\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}$
$\therefore\ \text{I}_2=-\frac{9}{2}\begin{Bmatrix}\frac{1}{2}\Big(\text{x}+\frac{3}{2}\Big)\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\\-\frac{1}{2}\Big(\frac{9}{2}\Big)^2\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\Big|\end{Bmatrix}+\text{C}$
$=-\frac{9}{4}\begin{Bmatrix}\Big(\frac{2\text{x}+3}{2}\Big)\sqrt{\text{x}^2+3\text{x}-18}-\Big(\frac{729}{4}\Big)\\\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|\end{Bmatrix}+\text{C}$
$=-\frac{9}{8}(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}+\frac{729}{16}\\\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|+\text{C}$
Thus,
$\text{I}=-\frac{1}{3}(\text{x}^2+3\text{x}-18)^{\frac{3}{2}}-\frac{9}{8}(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\\+\frac{729}{16}\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|+\text{C}$
View full question & answer→Question 165 Marks
Evaluate the following integrals:
$\int(2\text{x}+5)\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$
Answer$\text{I}=\int(2\text{x}+5)\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$Let $(2\text{x}+5)=\text{A}\frac{\text{d}}{\text{dx}}(10-4\text{x}-3\text{x}^2)+\text{B}$
$\Rightarrow(2\text{x}+5)=\text{A}(-4-6\text{x})+\text{B}$
$\Rightarrow(2\text{x}+5)=-6\text{A}\text{x}+(\text{B}-4\text{A})$
$\Rightarrow2=-6\text{A}\text{ and }(\text{B}-4\text{A})=5$
$\Rightarrow\text{A}=-\frac{1}{3}\text{ and }\text{B}=\frac{11}{3}$
$\Rightarrow(2\text{x}+5)=-\frac{1}{3}(-4-6\text{x})+\frac{11}{3}$
$\Rightarrow\text{I}=-\frac{1}{3}(-4-6\text{x})\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}\\+\frac{11}{3}\int\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$
Let $\text{I}=-\frac{1}{3}\text{I}_1+\frac{11}{3}\text{I}_2\ \dots(1)$
Now,
$\text{I}_1=\int(-4-6\text{x})\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$
Let $(10-4\text{x}-3\text{x}^2)=\text{t},\text{ or, }(-4-6\text{x})\text{dx = dt}$
$\Rightarrow\text{I}_1=\int\sqrt{\text{t}}\text{dt}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}+\text{c}_1$
$\Rightarrow\text{I}_1=\frac{2}{3}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}+\text{c}_1$
$\text{I}_2=\int\sqrt{(10-4\text{x}-3\text{x}^2)}\text{dx}$
$=\int\sqrt{3\Big(\frac{10}{3}-\frac{4}{3}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt3\int\sqrt{\Big(\frac{26}{9}-\frac{4}{9}-\frac{4}{3}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt3\int\sqrt{\Big[\Big(\frac{\sqrt{26}}{3}\Big)^2-\Big(\frac{4}{9}+\frac{4}{3}\text{x}-\text{x}^2\Big)\Big]}\text{dx}$
$=\sqrt3\int\sqrt{\Big[\Big(\frac{\sqrt{26}}{3}\Big)^2-\Big(\text{x}+\frac{2}{3}\Big)^2\Big]}\text{dx}$
$=\sqrt3\sin\Bigg(\frac{\text{x}+\frac{2}{3}}{\frac{\sqrt{26}}{3}}\Bigg)+\text{c}_2$
$=\sqrt3\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{c}_2$
Using (1), we get
$\text{I}=-\frac{1}{3}\times\frac{2}{3}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}\\+\frac{11}{3}\times\sqrt3\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{C}$
$\therefore\ \text{I}=-\frac{2}{9}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}+\frac{11\sqrt3}{3}\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{C}$
View full question & answer→Question 175 Marks
Evaluate the following integrals:
$\int(3\text{x}+1)\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$
Answer$\text{I}=\int(3\text{x}+1)\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$Let $(3\text{x}+1)=\text{A}\frac{\text{d}}{\text{dx}}(4-3\text{x}-2\text{x}^2)+\text{B}$
$\Rightarrow(3\text{x}+1)=\text{A}(-3-4\text{x})+\text{B}$
$\Rightarrow(3\text{x}+1)=-4\text{A}\text{x}+(\text{B}-3\text{A})$
$\Rightarrow3=-4\text{A}\text{ and }(\text{B}-3\text{A})=1$
$\Rightarrow\text{A}=-\frac{3}{4}\text{ and }\text{B}=-\frac{5}{4}$
$\Rightarrow(3\text{x}+1)=-\frac{3}{4}(-3-4\text{x})-\frac{5}{4}$
$\Rightarrow\text{I}=-\frac{3}{4}(-3-4\text{x})\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}\\-\frac{5}{4}\int\sqrt{4-3\text{x}-2\text{x}^2}$
Let $\text{I}=-\frac{3}{4}\text{I}_1-\frac{5}{4}\text{I}_2\ \dots(1)$
Now,
$\text{I}_1=\int(-3-4\text{x})\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$
Let $(4-3\text{x}-2\text{x}^2)=\text{t},\text{ or, }(-3-4\text{x})\text{dx = dt}$
$\Rightarrow\text{I}_1=\int\sqrt{\text{t}}\text{dt}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}+\text{c}_1$
$\Rightarrow\text{I}_1=\frac{2}{3}(4-3\text{x}-2\text{x}^2)^{\frac{3}{2}}+\text{c}_1$
$\text{I}_2=\int\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$
$=\int\sqrt{2\Big(2-\frac{3}{2}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt2\int\sqrt{\Big(\frac{17}{4}-\frac{9}{4}-\frac{3}{2}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt2\int\sqrt{\Big[\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\frac{9}{4}+\frac{3}{2}\text{x}+\text{x}^2\Big)\Big]}\text{dx}$
$=\sqrt2\int\sqrt{\Big[\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\text{x}+\frac{3}{2}\Big)^2\Big]}\text{dx}$
$=\sqrt2\sin\Bigg(\frac{\text{x}+\frac{3}{2}}{\frac{\sqrt{17}}{2}}\Bigg)+\text{c}_2$
$=\sqrt2\sin\Big(\frac{2\text{x}+3}{\sqrt{17}}\Big)+\text{c}_2$
Using (1), we get
$\text{I}=-\frac{3}{4}\times\frac{2}{3}(4-3\text{x}-2\text{x}^2)^{\frac{3}{2}}\\-\frac{5}{4}\times\sqrt2\sin\Big(\frac{2\text{x}+3}{\sqrt{17}}\Big)+\text{C}$
$\therefore\ \text{I}=-\frac{1}{2}(4-3\text{x}-2\text{x}^2)^{\frac{3}{2}}-\frac{5\sqrt2}{4}\sin\Big(\frac{2\text{x}+3}{\sqrt{17}}\Big)+\text{C}$
View full question & answer→Question 185 Marks
Evaluate the following intregals:
$\int\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}\ \text{dx}$
Answerwe have$\text{I}=\int\frac{\text{x}\text{dx}}{(\text{x}-1)(\text{x}+2)}$
Let $\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{(\text{x}-1)^2}+\frac{\text{C}}{\text{x}+2}$
$\Rightarrow\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{\text{A}(\text{x}-1)(\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}-1)^2}{(\text{x}-1)^2(\text{x}+2)}$
$\Rightarrow\text{x}=\text{A}(\text{x}^2+2\text{x}-\text{x}-2)+\text{B}(\text{x}+2)+\text{C}(\text{x}^2-2\text{x}+1)$
$\Rightarrow\text{x}=\text{A}(\text{x}^2+\text{x}-2)+\text{B}(\text{x}+2)+\text{C}(\text{x}^2-2\text{x}+1)$
$\Rightarrow\text{x}=(\text{A}+\text{C})\text{x}^2+(\text{A}+\text{B}-2\text{C})\text{x}+(-2\text{A}+2\text{B}+\text{C})$
A + C = 0 ...(1)
A + B - 2C = 1 ...(2)
-2A + 2B + C = 0 ...(3)
Solving(1), (2) and (3), we get
$\text{A}=\frac{2}{9},\text{B}=\frac{1}{3}\text{ and }\text{C}=-\frac{2}{9}$
$\therefore\frac{\text{x}}{(\text{x}-1)^2(\text{x}+2)}=\frac{2}{9(\text{x}-1)}+\frac{1}{3(\text{x}-1)^2}-\frac{2}{9(\text{x}+2)}$
$\Rightarrow\text{I}=\frac{2}{9}\int\frac{\text{dx}}{\text{x}-1}+\frac{1}{3}\int\frac{\text{dx}}{(\text{x}-1)^3}-\frac{2}{9}\int\frac{\text{dx}}{\text{x}+2}$
$=\frac{2}{9}\log|\text{x}-1|+\frac{1}{3}\times\Big(\frac{-1}{\text{x}-1}\Big)-\frac{2}{9}\log|\text{x}+2 |+\text{C}$
$=\frac{2}{9}\log\Big(\frac{\text{x}-1}{\text{x}+2}\Big)-\frac{1}{3(\text{x}-1)}+\text{C}$
View full question & answer→Question 195 Marks
Evaluate the following intregals: $\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$We express
$\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}=\frac{\text{x}^2}{\text{x}^4-4\text{x}^2+3\text{x}^2-12}$
$=\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}$
$=\frac{\text{A}}{\text{x}^2-4}+\frac{\text{B}}{\text{x}^2+3}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}^2+3)+\text{B}(\text{x}^2-4)$
Equating the coefficients of $x^2$ and constant, we get
$1 = A + B$ and $0 = 3A - 4B$ or
$\text{A}=\frac{4}{7}$ and $\text{B}=\frac{3}{7}$
$\therefore\text{I}=\int\bigg(\frac{\frac{4}{7}}{\text{x}^2-4}+\frac{\frac{3}{7}}{\text{x}^2+3}\bigg)\text{dx}$
$=\frac{4}{7}\int\frac{1}{\text{x}^2-4}\ \text{dx}+\frac{3}{7}\int\frac{1}{\text{x}^2+3}\text{dx}$
$=\frac{4}{7}\times\frac{1}{4}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
$=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
Hence, $\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$
$=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
View full question & answer→Question 205 Marks
Evaluate the following intregals:
$\int\frac{2\text{x}+5}{\sqrt{\text{x}^2+2\text{x}+5}}\text{dx}$
Answer Let $\text{I}=\int\frac{2\text{x}+5}{\sqrt{\text{x}^2+2\text{x}+5}}\text{dx}$let $2\text{x}+5=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x}+5)+\mu$
$=\mu(2\text{x}+2)+\mu$
$2\text{x}+5=(2\lambda)\text{x}+2\lambda+\mu$
Compairing the coefficient of like powewrs of x,
$2\lambda=2\ \Rightarrow\lambda=1$
$2\lambda+\mu=5\ \Rightarrow 2(1)+\mu=5$
$\Rightarrow\mu=3$
So, $\text{I}=\int\frac{(2\text{x}+2)+3}{\sqrt{\text{x}^2+2\text{x}+5}}\text{dx}$
$=\int\frac{(2\text{x}+3)}{\sqrt{\text{x}^2+2\text{x}+5}}\text{dx}+3\int\frac{1}{\sqrt{\text{x}^2+2\text{x}+(1)^2-(1)^2+5}}\text{dx}$
$\text{I}=\int\frac{2\text{x}+3}{\sqrt{\text{x}^2+2\text{x}+5}}+3\int\frac{1}{\sqrt{(\text{x}+1)^2+(2)^2}}\text{dx}$
$\text{I}=2\sqrt{\text{x}^22\text{x}+5}+3\log\big|\text{x}+1+{\sqrt{(\text{x}+1)^2+(2)^2}}\big|+\text{C}$ $\big[\text{since}, \int\frac{1}{\text{x}}\text{dx}=2\sqrt{\text{x}+\text{C},\int\frac{1}{\sqrt{\text{x}^2+\text{a}^2}}}\text{dx}=\log\big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}+\text{C}\big]$
$\text{I}=2\sqrt{\text{x}^2+2\text{x}+5}+3\log\big|\text{x}+1+\sqrt{\text{x}^2+2\text{x}+5}\big|+\text{C}$
View full question & answer→Question 215 Marks
Evaluate the following intregals: $\int\frac{\text{x}}{(\text{x}^2-\text{a}^2)(\text{x}^2-\text{b}^2)}\ \text{dx}$
AnswerWe have,$\text{I}=\int\frac{\text{x}\ \text{dx}}{(\text{x}^2-\text{a}^2)(\text{x}^2-\text{b}^2)}$
Putting $x^2= t$
$\Rightarrow 2\text{xdx} = \text{dt}$
$\Rightarrow\text{xdx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dt}}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}$
Let $\frac{1}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}=\frac{\text{A}}{\text{t}-\text{a}^2}+\frac{\text{B}}{\text{t}-\text{b}^2}$
$\Rightarrow\frac{1}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}=\frac{\text{A}(\text{t}-\text{b}^2)+\text{B}(\text{t}-\text{a}^2)}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}$
$\Rightarrow1=\text{A}(\text{t}-\text{b}^2)+\text{B}(\text{t}-\text{a}^2)$
Putting$ t = b^2$
$1=\text{A}(\text{a}^2-\text{b}^2)+\text{B}\times0$
$\Rightarrow\text{A}=\frac{1}{\text{a}^2-\text{b}^2}$
$\text{I}=\frac{1}{2}\int\frac{\text{dt}}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}\int\frac{}{}\frac{\text{dt}}{\text{t}-\text{a}^2}+\frac{1}{2((\text{b}^2-\text{a}^2)}\int\frac{\text{dt}}{\text{t}-\text{b}^2}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}[\log|\text{t}-\text{a}^2|+\frac{1}{2(\text{b}^2-\text{a}^2)}\log|\text{t}-\text{b}^2|+\text{C}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}[\log|\text{t}-\text{a}^2|-\log|\text{t}-\text{b}^2|]+\text{C}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}\Big[\log\Big|\frac{\text{t}-\text{a}^2}{\text{t}-\text{b}^2}\Big|\Big]+\text{C}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}\log\Big|\frac{\text{x}^2-\text{a}^2}{\text{x}^2-\text{b}^2}\Big|+\text{C}$
View full question & answer→Question 225 Marks
Evaluate the following intregals:
$\int\frac{\text{x}}{(\text{x}+1)(\text{x}^2+1)}\ \text{dx}$
Answerlet $\frac{\text{x}}{(\text{x}+1)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$$\Rightarrow\text{x}=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+1)$
Equation similar terms, we get
$\text{A}+\text{B}=0,\text{B}+\text{C}=1,\text{A}+\text{C}=0$
Solving, we get, $\text{A}=-\frac{1}{2},\text{B}=\frac{1}{2},\text{C}=\frac{1}{2}$
Thus,
$\text{I}=-\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{2}\int\frac{\text{x dx}}{\text{x}^2+1}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
$\text{I}=-\frac{1}{2}\log|\text{x}+1|+\frac{1}{4}\log|\text{x}^2+1|+\frac{1}{2}\tan^{-1}\text{x}+\text{c}$
View full question & answer→Question 235 Marks
Evaluvate the following intregals:
$\int\frac{8\cot\text{x}+1}{3\cot\text{x}+2}\ \text{dx}$
AnswerLet $\text{I}=\int\Big(\frac{8\cot\text{x}+1}{3\cot\text{x}+2}\Big)\ \text{dx}$$=\int\bigg(\frac{8\frac{\cos\text{x}}{\sin\text{x}}+1}{\frac{3\cos\text{x}}{\sin\text{x}}+2}\bigg)\text{dx}$
$=\int\Big(\frac{8\cos\text{x}+\sin\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
Now, Let $8\cos\text{x}+\sin\text{x}=\text{A}(3\cos\text{x}+2\sin\text{x})+\text{B}(-3\sin\text{x}+2\cos\text{x})\ \dots(1)$
$\Rightarrow8\cos\text{x}+\sin\text{x}+\sin\text{x}=3\text{A}\cos\text{x}+2\text{A}\sin\text{x}-3\text{B}\sin\text{x}+2\text{B}\cos\text{x}$
$\Rightarrow8\cos\text{x}+\sin\text{x}-(3\text{A}+2\text{B})\cos\text{x}+(2\text{A}-3\text{B})\sin\text{x}$
Equating the coefficient of like terms we get,
$2\text{A}-3\text{B}=1\ \dots(2)$
$3\text{A}+2\text{B}=8\ \dots(3)$
Solving eq (2) and eq (3) we get,
$\text{A}=2,\text{B}=1$
Thus, by substracting the value of A and B in eq (1) we get,
$\text{I}=\int\Big[\frac{2(3\cos\text{x}+2\sin\text{x})+1(-3\sin\text{x}+2\cos\text{x})}{(3\cos\text{x}+2\sin\text{x})}\Big]\text{dx}$
$=2\int\Big(\frac{3\cos\text{x}+2\sin\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}+\int\Big(\frac{-3\sin\text{x}+2\cos\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
$=2\int\text{dx}+\int\Big(\frac{-3\sin\text{x}+2\cos\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
Putting $3\cos\text{x}+2\sin\text{x}=\text{t}$
$\Rightarrow(-3\sin\text{x}+2\cos\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=2\int\text{dx}+\int\frac{1}{\text{t}}\text{dt}$
$=2\text{x}+\ln|\text{t}|+\text{C}$
$=2\text{x}+\ln|3\cos\text{x}+2\sin\text{x}|+\text{C}$
View full question & answer→Question 245 Marks
Evaluate the following integrals:
$\int\frac{\sin^3\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
AnswerLet I $=\int\frac{\sin^3\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$$\therefore\text{I}=\int\frac{\sin^2\text{x}\sin\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int\frac{(1-\cos^2\text{x})}{\sqrt{\cos\text{x}}}\sin\text{x dx}\ ...(1)$
Let $\cos\text{x}=\text{t}$ then, $\text{d}(\cos\text{x})=\text{dt}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\Rightarrow\sin\text{x dx}=-\text{dt}$
Putting $\cos\text{x}=\text{t}$ and $\sin\text{x dx}=-\text{dt}$ in equation (1), we get
$\text{I}=\int\frac{(1-\text{t}^2)}{\sqrt{\text{t}}}\times-\text{dt}$
$=\int\frac{\text{t}^2-1}{\sqrt{\text{t}}}\text{dt}$
$=\int\Bigg(\frac{\text{t}^2}{\text{t}^\frac{1}{2}}-\frac{1}{\text{t}^\frac{1}{2}}\Bigg)\text{dt}$
$=\int\Big(\text{t}^{2-\frac{1}{2}}-\text{t}^\frac{-1}{2}\Big)\text{dt}$
$=\int\Big(\text{t}^\frac{3}{2}-\text{t}^{\frac{-1}{2}}\Big)\text{dt}$
$=\frac{2}{5}\text{t}^\frac{5}{2}-2\text{t}^\frac{1}{2}+\text{C}$
$\therefore\text{I}=\frac{2}{5}\cos^\frac{5}{2}\text{x}-2\cos^\frac{1}{2}\text{x}+\text{C}$
$\therefore\text{I}=\frac{2}{5}\cos^\frac{5}{2}\text{x}-2\sqrt{\cos\text{x}}+\text{C}$
View full question & answer→Question 255 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{(2\text{x}+1)(\text{x}^2-1)}\text{ dx}$
AnswerWe have,$\text{I}=\int\frac{\text{x}^2+1}{(2\text{x}+1)(\text{x}^2-1)}$
$\text{I}=\int\frac{(\text{x}^2+1)\text{dx}}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}$
Let $\text{I}=\int\frac{(\text{x}^2+1)}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}=\frac{\text{A}}{2\text{x}+1}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow\int\frac{(\text{x}^2+1)}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}=\frac{\text{A}(\text{x}^2-1)+\text{B}(2\text{x}+1)(\text{x}+1)+\text{C}(2\text{x}+1)(\text{x}-1)}{(2\text{x}-1)(\text{x}-1)(\text{x}+1)}$
$\Rightarrow\text{x}^2+1=\text{A}(\text{x}^2-1)+\text{B}(2\text{x}+1)(\text{x}+1)\\+\text{C}(2\text{x}+1)(\text{x}-1)$
Putting x - 1 = 0
⇒ x = 1
1 + 1 = A × 0 + B × 0 + C (-2 + 1) (-1 - 1)
⇒ 2 = B (3) (2)
$\Rightarrow\text{B}=\frac{1}{3}$
Putting x + 1 = 0
⇒ x = -1
1 + 1 = A × 0 + B (-2 + 1)(-1 - 1)
⇒ 2 = C (-1) (-2)
⇒ C = 1
Putting 2x + 1 = 0
$\Rightarrow\text{x}=-\frac{1}{2}$
$\Big(-\frac{1}{2}\Big)^2+1=\text{A}\Big(\frac{1}{4}-1\Big)$
$\Rightarrow\frac{1}{4}+1=\text{A}\Big(-\frac{3}{4}\Big)$
$\Rightarrow\frac{5}{4}=\text{A}\Big(-\frac{3}{4}\Big)$
$\text{A}=-\frac{5}{3}$
$\therefore\text{I}=-\frac{5}{3}\int\frac{\text{dx}}{2\text{x}+1}+\frac{1}{3}\int\frac{\text{dx}}{\text{x}-1}+\int\frac{\text{dx}}{\text{x}+1}$
$=-\frac{5}{3}\times\frac{\log|2\text{x}+1|}{2}+\frac{1}{3}\log|\text{x}-1|+\log|\text{x}+1|+\text{C}$
$=-\frac{5}{6}\log|2\text{x}-1|+\frac{1}{3}\log|\text{x}-1|+\log|\text{x}+1|+\text{C}$
View full question & answer→Question 265 Marks
Evaluate the following intregals:
$\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}^2+2\text{x}^2+\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}^2+2\text{x}^2+\text{x}}\ \text{dx}$$=\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}(\text{x}+1)^2}\ \text{dx}$
Now,
Let $\frac{5\text{x}^2+20\text{x}+6}{\text{x}(\text{x}+1)^2}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{(\text{x}+1)^2}$
$\Rightarrow5\text{x}^2+20\text{x}+6=\text{A}(\text{x}+1)^2+\text{Bx}(\text{x}+1)+\text{Cx}$
Equating similar terms, we get,
A + B = 5, 2A + B + C = 20, A = 6
Solving, we get, B = -1, C = 9
Thus,
$\text{I}=\int\frac{6\text{dx}}{\text{x}}-1\int\frac{\text{dx}}{\text{x}+1}+9\int\frac{\text{dx}}{(\text{x}+1)^2}$
$\therefore\text{I}=6\log|\text{x}|-\log|\text{x}+1|-\frac{9}{\text{x}+1}+\text{C}$
View full question & answer→Question 275 Marks
$\int\frac{\text{x}^2}{\sqrt{\text{x}-1}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\sqrt{\text{x}-1}}\text{dx}$
Substituting x - 1 = t and dx = dt we get
$\text{I}=\int\frac{(\text{t}+1)^2}{\sqrt{\text{t}}}\text{dx}$
$=\int\frac{\text{t}^2+1+2\text{t}}{\sqrt{\text{t}}}\text{dt}$
$=\int\Big(\text{t}^\frac{3}{2}+\text{t}^\frac{-1}{2}+2\text{t}^\frac{-1}{2}\Big)\text{dt}$
$=\frac{2}{5}\text{t}^\frac{5}{2}+2\text{t}^\frac{1}{2}+\frac{4}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{6\text{t}^\frac{5}{2}+30\text{t}^\frac{1}{2}+20\text{t}^\frac{3}{2}}{15}+\text{C}$
$=\frac{2}{5}\text{t}^\frac{1}{2}\big(3\text{t}^2+15+10\text{t}\big)+\text{C}$
$=\frac{2}{5}\sqrt{\text{x}-1}\Big(3\big(\text{x}-1\big)^2+15+10(\text{x}-1)\Big)+\text{C}$
$=\frac{2}{5}\sqrt{\text{x}-1}\Big(3\big(\text{x}^2+1-2\text{x}\big)+15+10\text{x}-10\Big)+\text{C}$
$=\frac{2}{5}\sqrt{\text{x}-1}\big(3\text{x}^2+4\text{x}+8\big)+\text{C}$
$\therefore\ \text{I}=\frac{2}{5}\big(3\text{x}^2+4\text{x}+8\big)+\sqrt{\text{x}-1}+\text{C}$
View full question & answer→Question 285 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}\text{ dx}$
AnswerLet $\text{I}=\int\Big(\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}\Big)\text{dx}$ Dividing Numerator by Denominator
$\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}=1+\Big(\frac{5\text{x}-5}{\text{x}^2-5\text{x}+6}\Big)\ ....(1)$ Also $\frac{5\text{x}-5}{\text{x}^2-5\text{x}+6}=\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}$ Let $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}}{\text{x}-2}+\frac{\text{B}}{\text{x}-3}$ $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}(\text{x}-3)+\text{B}(\text{x}-2)}{(\text{x}-2)(\text{x}-3)}$ $\Rightarrow5\text{x}-5=\text{A}(\text{x}-3)+\text{B}(\text{x}-2)$ Let $\text{x}=3$ $5\times3-5=\text{}A\times0+\text{B}(3-2)$ $10=\text{B}$ Let $\text{x}=2$ $5\times2-5=\text{}A(2-3)+\text{B}\times0$ $\text{A}=-5$ $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{-5}{\text{x}-2}+\frac{10}{\text{x}-3}\ ... ..(2)$ From (1) and (2) $\text{I}=\int\text{dx}-5\int\frac{\text{dx}}{\text{x}-2}+10\int\frac{\text{dx}}{\text{x}-3}$ $=\text{x}-5\log|\text{x}-2|+10\log|\text{x}-3|+\text{C}$ View full question & answer→Question 295 Marks
Evaluate the following integrals:$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-\cos^2\text{x}-4\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-\cos^2\text{x}-4\sin\text{x}}\text{ dx}$ $\therefore\ \text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-(1-\sin^2\text{x})-4\sin\text{x}}\text{ dx}$ $\Rightarrow\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-1+\sin^2\text{x}-4\sin\text{x}}\text{ dx}$ Substitute $\sin\text{x}=\text{t}$ $\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$ Thus, $\text{I}=\int\frac{(3\text{t}-2)}{4+\text{t}^2-4\text{t}}\text{ dt}$ $\text{I}=\int\frac{(3\text{t}-2)}{\text{t}^2-4\text{t}+4}\text{ dt}$ $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ Now let us separate the integrand into the simplest form using partial fractions. $\frac{(3\text{t}-2)}{(\text{t}-2)^2}=\frac{\text{A}}{(\text{t}-2)}+\frac{\text{B}}{(\text{t}-2)^2}$ $=\frac{\text{A}(\text{t}-2)+\text{B}}{(\text{t}-2)^2}$ $=\frac{\text{At}-2\text{A}+\text{B}}{(\text{t}-2)^2}$ $\Rightarrow3\text{t}-2=\text{At}-2\text{A}+\text{B}$ Comparing the coefficients, we have, $\text{A}=3$and
$-2\text{A}+\text{B}=-2$ Substituting the value of A = 3 in the above equation, we have, $\Rightarrow-2\times3+\text{B}=-2$ $\Rightarrow-6+\text{B}=-2$ $\Rightarrow\text{B}=6-2$ $\Rightarrow\text{B}=4$Thus, $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ becomes,
$\text{I}=\int\frac{3}{(\text{t}-2)^2}\text{ dt}+\int\frac{4}{(\text{t}-2)^2}\text{ dt}$ $=3\log|\text{t}-2|-4\Big(\frac{1}{\text{t}-2}\Big)+\text{C}$ $=3\log|2-\text{t}|+4\Big(\frac{1}{2-\text{t}}\Big)+\text{C}$ Now, substituting $\text{t}=\sin\text{x},$ we have, $=3\log|2-\sin\text{x}|+4\Big(\frac{1}{2-\sin\text{x}}\Big)+\text{C}$
View full question & answer→Question 305 Marks
Evaluate the following integrals:$\int\frac{(1-\text{x}^2)}{\text{x}(1-2\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1-\text{x}^2}{\text{x}(1-2\text{x})}\text{ dx}$ $=\int\frac{1-\text{x}^2}{\text{x}-2\text{x}^2}\text{ dx}$ $=\int\frac{1-\text{x}^2}{2\text{x}^2-\text{x}}\text{ dx}$ $=\int\bigg[\frac{1}{2}+\frac{\frac{\text{x}}{2}-1}{2\text{x}^2-\text{x}}\bigg]\text{dx}$ $\text{I}=\frac{1}{2}\text{x}+\int\frac{\frac{\text{x}}{2}-1}{2\text{x}^2-\text{x}}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{2\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$ Let $\frac{\text{x}}{2}-1=\lambda\frac{\text{d}}{\text{dx}}\big(2\text{x}^2-\text{x}\big)+\mu$ $=\lambda(4\text{x}-1)+\mu$ $\frac{\text{x}}{2}-1=(4 \lambda)\text{x}-\lambda+\mu$Comparing the coefficients of like powers of x,
$\frac{1}{2}=4\lambda\Rightarrow\lambda=\frac{1}{8}$ $-\lambda+\mu=-1\Rightarrow-\Big(\frac{1}{8}\Big)+\mu=-1$ $\mu=-\frac{7}{8}$ So, $\text{I}_1=\int\frac{\frac{1}{8}(4\text{x}-1)-\frac{7}{8}}{2\text{x}^2-\text{x}}\text{ dx}$ $\text{I}=\frac{1}{8}\int\frac{4\text{x}-1}{2\text{x}^2-\text{x}}\text{ dx}-\frac{7}{8}\int\frac{1}{2\big(\text{x}^2-\frac{\text{x}}{2}\big)}\text{ dx}$ $\text{I}=\frac{1}{8}\int\frac{4\text{x}-1}{2\text{x}^2-\text{x}}\text{ dx}-\frac{7}{16}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{4}\big)+\big(\frac{1}{4}\big)^2-\big(\frac{1}{4}\big)^2}\text{ dx}$ $\text{I}=\frac{1}{8}\int\frac{4\text{x}-1}{2\text{x}^2-\text{x}}\text{ dx}-\frac{7}{16}\int\frac{1}{\big(\text{x}-\frac{1}{4}\big)^2-\big(\frac{1}{4}\big)^2}\text{ dx}$ $\text{I}=\frac{1}{8}\log\big|2\text{x}^2+\text{x}\big|-\frac{7}{16}\times\frac{1}{2\big(\frac{1}{24}\big)}\log\bigg|\frac{\text{x}-\frac{1}{4}-\frac{1}{4}}{\text{x}-\frac{1}{4}+\frac{1}{4}}\bigg|+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}=\frac{1}{8}\log|\text{x}|+\frac{1}{8}\log|2\text{x}-1|-\frac{7}{8}\log|1-2\text{x}|+\frac{7}{8}\log2+\frac{7}{8}\log|\text{x}|+\text{C}_2$ $\text{I}_1=\log\big|\text{x}\big|-\frac{3}{4}\log|1-2\text{x}|+\text{C}_3\ ....(2)$ $\Big[\text{Say},\text{C}_3=\text{C}_2+\frac{7}{8}\log2\Big]$ Using equation (1) and (2) $\text{I}=\frac{1}{2}\text{x}+\log|\text{x}|-\frac{3}{4}\log|1-2\text{x}|+\text{C}$
View full question & answer→Question 315 Marks
Evaluate the following integrals:$\int\frac{\sin^{-1}\text{x}}{\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin^{-1}\text{x}}{\text{x}^2}\text{dx}$
$=\int\Big(\frac{1}{\text{x}^2}\Big)(\sin^{-1}\text{x})\text{dx}$
$\text{I}=\Big[\sin^{-1}\text{x}\int\frac{1}{\text{x}^2}\text{dx}-\int\Big(\frac{1}{\sqrt{1-\text{x}^2}}\int\frac{1}{\text{x}^2}\text{dx}\Big)\text{dx}\Big]$
$=\sin^{-1}\text{x}\big(-\frac{1}{\text{x}}\big)-\int\frac{1}{\sqrt{1-\text{x}^2}}\Big(-\frac{1}{\text{x}}\Big)\text{dx}$
$\text{I} =-\frac{1}{\text{x}}\sin^{-1}\text{x}+\int\frac{1}{\text{x}\sqrt{1-\text{x}^2}}\text{dx}$
$\text{I}=-\frac{1}{\text{x}}\sin^{-1}\text{x}+\text{I}_1 \dots(1)$
Where,
$\text{I}_1=\int\frac{1}{\text{x}\sqrt{1-\text{x}^2}}\text{dx}$
Let $1-\text{x}^2=\text{t}^2$
$-2\text{x dx}=2\text{t dt}$
$\text{I}_1=\int\frac{\text{x}}{\text{x}^2\sqrt{1-\text{x}^2}}\text{dx}$
$=-\int\frac{\text{tdt}}{(1-\text{t}^2)\sqrt{\text{t}}}$
$=-\int\frac{\text{dt}}{(1-\text{t}^2)}$
$=\int\frac{1}{\text{t}^2-1}\text{dt}$
$=\frac{1}{2}\log\Big|\frac{\text{t}-1}{\text{t+1}}\Big|$
$=\frac{1}2{\log}\Big|\frac{\sqrt{1-\text{x}^2}-1}{\sqrt{1-\text{x}^2}+1}\Big|+\text{C}_1$
Now
$\text{I}=-\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{1}{2}\log\bigg|\Big(\frac{\sqrt{1-\text{x}^2}-1}{\sqrt{1-\text{x}^2+1}}\Big)\Big(\frac{\sqrt{1-\text{x}^2}-1}{\sqrt{1-\text{x}^2-1}}\Big)\bigg|+\text{C}$
$=-\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{1}2{}\log\bigg|\frac{(\sqrt{1-\text{x}^2}-1)^2}{1-\text{x}^2-1}\bigg|+\text{C}$
$=-\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{1}{2}\log\bigg|\frac{(\sqrt{1-\text{x}^2}-1)^2}{-\text{x}^2}\bigg|+\text{C}$
$=-\frac{\sin^{-1}\text{x}}{\text{x}}+\log\bigg|\frac{\sqrt{1-\text{x}^2}-1}{-\text{x}}\bigg|+\text{C}$
$\text{I}=-\frac{\sin^{-1}\text{x}}{\text{x}}+\log\bigg|\frac{1-\sqrt{1-\text{x}}^2}{\text{x}}\bigg|+\text{C}$
View full question & answer→Question 325 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$ $=\int\Big[1+\frac{2\text{x}+1}{\text{x}^2-\text{x}}\Big]\text{dx}$ $=\text{x}+\int\frac{2\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{2\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$ Let $2\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2-\text{x}\big)+\mu$ $=\lambda(2\text{x}-1)+\mu$ $2\text{x}+1=(2\lambda)\text{x}-\lambda+\mu$Comparing the coefficients of like powers of x,
$2=2\lambda\Rightarrow\lambda=1$ $-\lambda+\mu=1\Rightarrow\mu=2$ So, $\text{I}_1=\int\frac{(2\text{x}-1)+2}{\text{x}^2-\text{x}}\text{ dx}$ $\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}}\text{ dx}+2\int\frac{1}{\text{x}^2-\text{x}}\text{ dx}$ $\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}}\text{ dx}+2\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{ dx}$ $\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}}\text{ dx}+2\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{ dx}$ $\text{I}=\log\big|\text{x}^2+\text{x}\big|+2\times\frac{1}{2\big(\frac{1}{2}\big)}\log\bigg|\frac{\text{x}-\frac{1}{2}-\frac{1}{2}}{\text{x}-\frac{1}{2}+\frac{1}{2}}\bigg|+\text{C}_1$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}_1=\log\big|\text{x}^2+\text{x}\big|+2\log\Big|\frac{\text{x}-1}{\text{x}}\Big|+\text{C}_2\ ....(2)$ Using equation (1) and (2) $\text{I}=\text{x}+\log\big|\text{x}^2+\text{x}\big|+2\log\Big|\frac{\text{x}-1}{\text{x}}\Big|+\text{C}$
View full question & answer→Question 335 Marks
Evaluate the following integrals:$\int\frac{\text{x}-1}{3\text{x}^2-4\text{x}+3}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{dx}$
$=\int\begin{Bmatrix}1-\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\end{Bmatrix}\text{dx}$
$\text{I}=\text{x}-\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{dx}+\text{c}_1\dots\text{(i})$
Let $\text{I}_1=\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{dx}$
Let $7\text{x}+10=\lambda\frac{\text{d}}{\text{dx}}\Big(\text{x}^2+7\text{x}+10{}\Big)+\mu$
$=\lambda(2\text{x}+7)+\mu$
$7\text{x}+10=(2\lambda)\text{x}+7\lambda+\mu$
Comparing the coefficients of like powers of x,
$7=2\lambda$
$\Rightarrow\lambda=\frac72$
$7\lambda+\mu=10$
$\Rightarrow7\Big(\frac72\Big)+\mu=10$
$\mu=-\frac{29}{2}$
View full question & answer→Question 345 Marks
Evaluate the following integrals:$\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$ Rewriting the numerator we have, $5\text{x}-2=\text{A}\frac{\text{d}}{\text{dx}}\big(1+2\text{x}+3\text{x}^2\big)+\text{B}$ $\Rightarrow5\text{x}-2=\text{A}(2+6\text{x})+\text{B}$ $\Rightarrow5\text{x}-2=6\text{xa}+2\text{A}+\text{B}$ Comparing the coefficient, we have, $6\text{A}=5$ and $2\text{A}+\text{B}=-2$ $\Rightarrow\text{A}=\frac{5}{6}$ Substituting the value of A in 2A + B = -2, we have, $2\times\frac{5}{6}+\text{B}=-2$ $\Rightarrow\frac{10}{6}+\text{B}=-2$ $\Rightarrow\text{B}=-2-\frac{10}{6}$ $\Rightarrow\text{B}=\frac{-12-10}{6}$ $\Rightarrow\text{B}=\frac{-22}{6}$ $\Rightarrow\text{B}=\frac{-11}{3}$ $5\text{x}-2=\frac{5}{6}(2+6\text{x})-\frac{11}{3}$Thus, $\text{I}=\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$ becomes,
$\text{I}=\int\frac{\big[\frac{5}{6}(2+6\text{x})-\frac{11}{3}\big]}{3\text{x}^2+2\text{x}+1}\text{ dx}$ $=\frac{5}{6}\int\frac{(2+6\text{x})}{3\text{x}^2+2\text{x}+1}\text{ dx}-\frac{11}{3}\int\frac{\text{dx}}{3\text{x}^2+2\text{x}+1}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{3\times3}\int\frac{\text{dx}}{\text{x}^2+\frac{2}{3}\text{x}+\frac{1}{3}}+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\int\frac{\text{dx}}{\text{x}^2+\frac{2}{3}\text{x}+\big(\frac{4}{3}\big)^2+\frac{1}{3}-\big(\frac{4}{3}\big)^2}+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\int\frac{\text{dx}}{\big(\text{x}+\frac{1}{3}\big)^2+\Big(\frac{\sqrt2}{3}\Big)^2}+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\times\frac{1}{\frac{\sqrt2}{3}}\tan^{-1}\Bigg[\frac{\big(\text{x}+\frac{1}{3}\big)}{\frac{\sqrt2}{3}}\Bigg]+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\times\frac{3}{\sqrt2}\tan^{-1}\Bigg[\frac{\big(\frac{3\text{x}+1}{3}\big)}{\frac{\sqrt2}{3}}\Bigg]+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{3\sqrt2}\tan^{-1}\Big[\frac{3\text{x}+1}{\sqrt2}\Big]+\text{C}$
View full question & answer→Question 355 Marks
Evaluate the following integrals:$\int\frac{\text{x}^3-3\text{x}}{\text{x}^4+2\text{x}^2-4}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}^3-3\text{x}}{\text{x}^4+2\text{x}^2-4}\text{ dx}$ $=\int\frac{\text{x}(\text{x}^2-3)}{\text{x}^4+2\text{x}^2-4}\text{ dx}$ Let $\text{x}^2=\text{t},$ or, $2\text{x}\text{ dx}=\text{dt}$ $\text{I}=\frac{1}{2}\int\frac{(\text{t}-3)}{\text{t}^2+2\text{t}-4}\text{ dt}$ $=\frac{1}{4}\int\frac{2\text{t}-6}{\text{t}^2+2\text{t}-4}\text{ dt}$$=\frac{1}{4}\int\frac{2\text{t}+2-8}{\text{t}^2+2\text{t}-4}\text{ dt}$
$=\frac{1}{4}\int\Big(\frac{2\text{t}+2}{\text{t}^2+2\text{t}-4}-\frac{8}{\text{t}^2+2\text{t}-4}\Big)\text{dt}$
$=\frac{1}{4}\Big(\int\frac{2\text{t}+2}{\text{t}^2+2\text{t}-4}\text{ dt}-\int\frac{8}{\text{t}^2+2\text{t}-4}\text{ dt}\Big)$ $\Rightarrow\text{I}=\frac{1}{4}(\text{I}_1+\text{I}_2)\ ....(1)$ Now, $\text{I}_1=\int\frac{2\text{t}+2}{\text{t}^2+2\text{t}-4}\text{ dt}$ Let $\text{t}^2+2\text{t}-4=\text{u}$ or, $(2\text{t}+2)\text{dt}=\text{du}$ $\Rightarrow\text{I}_1=\int\frac{1}{\text{u}}\text{ du}=\int|\text{u}|+\text{C}_1$ $\Rightarrow\text{I}_1=\ln|\text{t}^2+2\text{t}-4|+\text{C}_1$ $\therefore\ \text{I}_1=\ln|\text{x}^4+2\text{x}^2-4|+\text{C}_1$ Now, $\text{I}_2=\int\frac{-8}{(\text{t}+1)^2-5}\text{ dt}$ $\Rightarrow\text{I}_2=\int\frac{8}{\big(\sqrt5\big)^2-(\text{t}+1)^2}\text{ dt}$ $\therefore\ \text{I}_2=\frac{8}{2\sqrt5}\ln\Big|\frac{\sqrt5+\text{x}^2+1}{\sqrt5-\text{x}^2-1}\Big|+\text{C}_2$ So, from (1), we get $\text{I}=\frac{1}{4}\Big[\ln|\text{x}^4+2\text{x}^2-4|+\frac{4}{\sqrt5}\ln\Big|\frac{\sqrt5+\text{x}^2+1}{\sqrt5-\text{x}^2-1}\Big|\Big]+\text{C}$ $\therefore\ \text{I}=\frac{1}{4}\ln|\text{x}^4+2\text{x}^2-4|+\frac{1}{\sqrt5}\ln\Big|\frac{\sqrt5+\text{x}^2+1}{\sqrt5-\text{x}^2-1}\Big|+\text{C}$
View full question & answer→Question 365 Marks
Evaluate the following integrals:$\int\frac{(\text{x}-1)^2}{\text{x}^2+2\text{x}+2}\text{ dx}$
AnswerLet $\int\frac{(\text{x}-1)^2}{\text{x}^2+2\text{x}+2}\text{ dx}$ $=\int\Big(\frac{\text{x}^2-2\text{x}+1}{\text{x}^2-2\text{x}+2}\Big)\text{dx}$ Here,
Therefore,
$\frac{\text{x}^2-2\text{x}+1}{\text{x}^2+2\text{x}+2}=1-\frac{(4\text{x}+1)}{\text{x}^2+2\text{x}+2}\ ....(1)$ Let $4\text{x}+1=\text{A}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+2\text{x}+2\big)+\text{B}$ $4\text{x}+1=\text{A}(2\text{x}+2)+\text{B}$ $4\text{x}+1=(2\text{A})\text{x}+2\text{A}+\text{B}$ Equating Cofficients of like terms $2\text{A}=4$ $\text{A}=2$ $2\text{A}+\text{B}=1$ $2\times2+\text{B}=1$ $\text{B}=-3$ $\int\Big(\frac{\text{x}^2-2\text{x}+1}{\text{x}^2-2\text{x}+2}\Big)\text{dx}$ $=\int\text{dx}-2\int\frac{(2\text{x}+2)}{\text{x}^2+2\text{x}+2}\text{ dx}+3\int\frac{\text{ dx}}{\text{x}^2+2\text{x}+2}$ $=\int\text{dx}-2\int\frac{(2\text{x}+2)}{\text{x}^2+2\text{x}+2}\text{ dx}+3\int\frac{\text{dx}}{(\text{x}+1)^2+1^2}$ $=\text{x}-2\log\big|\text{x}^2+2\text{x}+2\big|+\frac{3}{1}\tan^{-1}\Big(\frac{\text{x}+1}{1}\Big)+\text{C}$ $=\text{x}-2\log\big|\text{x}^2+2\text{x}+2\big|+3\tan^{-1}(\text{x}+1)+\text{C}$ View full question & answer→Question 375 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2+\text{x}+1}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2+\text{x}+1}\text{dx}$ $=\int\begin{bmatrix}1+\frac{\text{2x}}{\text{x}^2-\text{x}+1}\end{bmatrix}\text{dx}$ $\text{I}=\text{x}+\int\frac{2\text{x}}{\text{x}^2-\text{x}+1}\text{ dx}+\text{C}_1\ ....(1)$ Let $\text{I}=\int\frac{2\text{x}}{\text{x}^2-\text{x}+1}\text{ dx}$ Let $2\text{x}=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2-\text{x}+1\big)+\mu$ $=\lambda(2\text{x}-1)+\mu$ $2\text{x}=(2\lambda)\text{x}-\lambda+\mu$Comparing the coefficients of like powers of x,
$2=2\lambda$ $\Rightarrow\lambda=1$ $-\lambda+\mu=0$ $\Rightarrow-1+\mu=0$ $\mu=1$ So, $\text{I}_1=\int\frac{(2\text{x}-1)+1}{\text{x}^2-\text{x}+1}\text{ dx}$ $=\int\frac{(2\text{x}-1)}{\text{x}^2-\text{x}+1}\text{ dx}+\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+1}\text{ dx}$ $\text{I}_1=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}+1}\text{ dx+}\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\Big(\frac{\sqrt3}{2}\Big)^2}\text{ dx}$ $=\log\big|\text{x}^2-\text{x}+1\big|+\frac{2}{\sqrt3}\tan^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{\sqrt3}{2}}\bigg)+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$ $=\log\big|\text{x}^2-\text{x}+1\big|+\frac{2}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}-1}{\sqrt3}\Big)+\text{C}_2\ .....(2)$ Using equation (1) and (2) $\text{I}=\text{x}+\log\big|\text{x}^2-\text{x}+1\big|+\frac{2}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}-1}{\sqrt3}\Big)+\text{C}$
View full question & answer→Question 385 Marks
Evaluate the following integrals:$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-\cos^2\text{x}-7\sin\text{x}}\text{ dx}$
Answer$\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-\cos^2\text{x}-7\sin\text{x}}\text{ dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-(1-\sin^2\text{x})-7\sin\text{x}}\text{ dx}$ $\big(\because\ \cos^2\text{x}=1-\sin^2\text{x}\big)$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin^2\text{x}-7\sin\text{x}+12}\text{ dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin^2\text{x}-4\sin\text{x}-3\sin\text{x}+12}\text{ dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin\text{x}(\sin\text{x}-4)-3(\sin\text{x}-4)}\text{dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{(\sin\text{x}-3)(\sin\text{x}-4)}\text{ dx}$Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-3)(\text{t}-4)}\text{ dt}$
Using partial fraction, we get $\frac{(3\text{t}-2)}{(\text{t}-3)(\text{t}-4)}=\frac{\text{A}}{(\text{t}-3)}+\frac{\text{B}}{(\text{t}-4)}$ $=\frac{\text{A}(\text{t}-4)+\text{B}(\text{t}-3)}{(\text{t}-3)(\text{t}-4)}$ $\Rightarrow3\text{t}-2=(\text{A}+\text{B})\text{t}-4\text{A}-3\text{B}$ Comparing coefficients, we get A = -7 and B = 10 So, $\text{I}=-7\int\frac{1}{(\text{t}-3)}\text{ dt}+10\int\frac{1}{(\text{t}-4)}\text{ dt}$$\Rightarrow\text{I}=-7\ln|\text{t}-3|+10\ln|\text{t}-4|+\text{C}$
$\therefore\ \text{I}=-7\ln|\sin\text{x}-3|+10\ln|\sin\text{x}-4|+\text{C}$
View full question & answer→Question 395 Marks
Evaluate the following integrals:$\int\frac{\text{x}-3}{\text{x}^2+2\text{x}-4}\text{ dx}$
Answer$\int\Big(\frac{\text{x}-3}{\text{x}^2+2\text{x}-4}\Big)\text{dx}$
$\text{x}-3=\text{A}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+2\text{x}-4\big)+\text{B}$
$\text{x}-3=\text{A}(2\text{x}+2)+\text{B}$
$\text{x}-3=(2\text{A})\text{x}+2\text{A}+\text{B}$
Comparing the coefficients of like power of x,
$2\text{A}=1$
$\text{A}=\frac{1}{2}$
$2\text{A}+\text{B}=-3$
$2\times\frac{1}{2}+\text{B}=-3$
$\text{B}=-4$
Now, $\int\Big(\frac{\text{x}-3}{\text{x}^2+2\text{x}-4}\Big)\text{dx}$
$=\int\bigg(\frac{\frac{1}{2}(2\text{x}+2)-4}{\text{x}^2+2\text{x}-4}\bigg)\text{dx}$
$=\frac{1}{2}\int\frac{(2\text{x}+2)\text{dx}}{(\text{x}^2+2\text{x}-4)}-4\int\frac{\text{dx}}{\text{x}^2+2\text{x}+1-1-4}$
$=\frac{1}{2}\int\frac{(2\text{x}+2)\text{dx}}{(\text{x}^2+2\text{x}-4)}-4\int\frac{\text{dx}}{(\text{x}+1)^2-(\sqrt5)^2}$
$=\frac{1}{2}\log\big|\text{x}^2+2\text{x}+4\big|-\frac{4}{2\sqrt5}\log\Big|\frac{\text{x}+1-\sqrt5}{\text{x}+1+\sqrt5}\Big|+\text{C}$
$=\frac{1}{2}\log\big|\text{x}^2+2\text{x}+4\big|-\frac{2}{\sqrt5}\log\Big|\frac{\text{x}+1-\sqrt5}{\text{x}+1+\sqrt5}\Big|+\text{C}$
View full question & answer→Question 405 Marks
Evaluate the following integrals:$\int\frac{\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}}{\text{x}^2+\text{x}+3}\text{ dx}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{x}+3\big)+\mu$
$\text{x}+1=\lambda(2\text{x}+1)+\mu$
$\text{x}+1=(2\lambda)\text{x}+(\lambda+\mu)$
Comparing the coefficients of like powers of x,
$2\lambda=1$
$\Rightarrow\lambda=\frac{1}{2}$
$\lambda+\mu=1$
$\Rightarrow\Big(\frac{1}{2}\Big)+\mu=0$
$\mu=\frac{1}{2}$
So, $\text{I}=\int\frac{\frac{1}{2}(2\text{x}+1)+\frac{1}{2}}{\text{x}^2+\text{x}+3}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}+\frac{1}{2}\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+3}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}+\frac{1}{2}\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2+\big(\frac{11}{4}\big)}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+1}{\text{x}^2+\text{x}+3}\text{ dx}+\frac{1}{2}\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2+\big(\frac{\sqrt{11}}{2}\big)^2}\text{ dx}$
$\text{I}=\frac{1}{2}\log\big|\text{x}^2+\text{x}+3\big|+\frac{1}{2}\times\frac{1}{\Big(\frac{\sqrt{11}}{2}\Big)}\tan^{-1}\bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt{11}}{2}}\bigg)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\frac{1}{2}\log\big|\text{x}^2+\text{x}+3\big|+\frac{1}{\sqrt{11}}\tan^{-1}\Big(\frac{2\text{x}+1}{\sqrt{11}}\Big)+\text{C}$
View full question & answer→Question 415 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{ dx}$ $=\int\Big\{1-\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\Big\}\text{dx}$ $=\text{x}-\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{ dx}$ Let $7\text{x}+10=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+7\text{x}+10\big)+\mu$ $=\lambda(2\text{x}+7)+\mu$ $7\text{x}+10=(2\lambda)\text{x}+7\lambda+\mu$Comparing the coefficients of like powers of x,
$7=2\lambda\Rightarrow\lambda=\frac{7}{2}$ $7\lambda+\mu=10\Rightarrow7\Big(\frac{7}{2}\Big)+\mu=10$ $\mu=-\frac{29}{2}$ So, $\text{I}_1=\int\frac{\frac{7}{2}(2\text{x}+7)-\frac{29}{2}}{\text{x}^2+7\text{x}+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{(2\text{x}+7)}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{7}{2}\big)+\big(\frac{7}{2}\big)^2-\big(\frac{7}{2}\big)^2+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{2\text{x}+7}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\big(\text{x}+\frac{7}{2}\big)^2-\big(\frac{3}{2}\big)^2}\text{ dx}$ $\text{I}_1=\frac{2}{7}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{2}\times\frac{1}{2\big(\frac{3}{2}\big)}\log\bigg|\frac{\text{x}+\frac{7}{2}-\frac{3}{2}}{\text{x}+\frac{7}{2}+\frac{3}{2}}\bigg|+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}_1=\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}_2\ ....(2)$ Using equation (1) and (2) $\text{I}=\text{x}-\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|+\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}$
View full question & answer→Question 425 Marks
Evaluate the following integrals:$\int\frac{2\text{x}-3}{\text{x}^2+6\text{x}+13}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2\text{x}-3}{\text{x}^2+6\text{x}+13}\text{ dx}$
Let $2\text{x}-3=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+6\text{x}+13\big)+\mu$
$=\lambda(2\text{x}+6)+\mu$
$2\text{x}-3=(2\lambda)\text{x}+(6\lambda+\mu)$
Comparing the coefficients of like powers of x,
$2\lambda=2\Rightarrow\lambda=1$
$6\lambda+\mu=-3\Rightarrow6(1)+\mu=-3$
$\mu=-9$
So, $\text{I}=\int\frac{1(2\text{x}+6)-9}{\text{x}^2+6\text{x}+13}\text{ dx}$
$\text{I}=\int\frac{2\text{x}+6}{\text{x}^2+6\text{x}+13}\text{ dx}-9\int\frac{1}{\text{x}^2+2\text{x}(3)+(3)^2-(3)^2+13}\text{ dx}$
$\text{I}=\int\frac{2\text{x}+6}{\text{x}^2+6\text{x}+13}\text{ dx}-9\int\frac{1}{(\text{x}+3)^2+(2)^2}\text{ dx}$
$\text{I}=\log\big|\text{x}^2+6\text{x}+13\big|-9\times\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}+3}{2}\Big)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\log\big|\text{x}^2+6\text{x}+13\big|-\frac{9}{2}\tan^{-1}\Big(\frac{\text{x}+3}{2}\Big)+\text{C}$
View full question & answer→Question 435 Marks
Evaluate the following integrals:$\int\frac{\text{ax}^3+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{ax}^3+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$ Let $\text{ax}^3+\text{bx}=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^4+\text{c}^2\big)+\mu$ $\text{ax}^3+\text{bx}=\lambda\big(4\text{x}^3\big)+\mu$ Comparing the coefficients of like powers of x, $4\lambda=\text{a}\Rightarrow\lambda=\frac{\text{a}}{4}$ $\mu=0\Rightarrow\mu=0$ So, $\text{I}=\int\frac{\frac{\text{a}}{4}(4\text{x}^3)+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\int\frac{4\text{x}^3}{\text{x}^4+\text{c}^2}\text{ dx}+\text{b}\int\frac{\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\int\frac{4\text{x}^3}{\text{x}^4+\text{c}^2}\text{ dx}+\frac{\text{b}}{2}\int\frac{2\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\log\big|\text{x}^4+\text{c}^2\big|+\frac{\text{b}}{2}\text{I}_1\ ....(1)$Now,
$\text{I}_1=\int\frac{2\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ Put $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}\text{ dx}=\text{dt}$ $\text{I}_1=\int\frac{1}{(\text{t})^2+\text{c}^2}\text{ dx}$ $\text{I}_1=\frac{1}{\text{c}}\tan^{-1}\Big(\frac{\text{t}}{\text{c}}\Big)+\text{C}_1\ ....(2)$ Using equation (2) in equation (1), $\text{I}=\frac{\text{a}}{4}\log\big|\text{x}^4+\text{c}^4\big|+\frac{\text{b}}{2\text{c}}\tan^{-1}\Big(\frac{\text{x}^2}{\text{c}}\Big)+\text{K}$ K = Integration constant.
View full question & answer→Question 445 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2(\text{x}^4+4)}{\text{x}^2+4}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2(\text{x}^4+4)}{\text{x}^2+4}\text{ dx}$
$=\int\frac{\text{x}^6+4\text{x}^2}{(\text{x}^2+4)}\text{ dx}$
$=\int\Big[\text{x}^4-4\text{x}^2+20-\frac{80}{\text{x}^2+4}\Big]\text{dx}$
$\text{I}=\frac{\text{x}^5}{5}-\frac{4\text{x}^3}{3}+20\text{x}-80\int\frac{1}{\text{x}^2+4}\text{ dx}+\text{C}_1\ ....(1)$
Let $\text{I}_1=\int\frac{1}{\text{x}^2+4}\text{ dx}$
$\text{I}_1=\int\frac{1}{\text{x}^2+(2)^2}\text{ dx}$
$\text{I}_1=\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}_2\ ....(2)$ $\Big[\text{Since},\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
Using equation (1) and (2)
$\text{I}=\frac{\text{x}^5}{5}-\frac{4\text{x}^3}{3}+20\text{x}-\frac{80}{2}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$
$\text{I}=\frac{\text{x}^5}{5}-\frac{4\text{x}^3}{3}+20\text{x}-40\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$
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Evaluate the following intregals:$\int\frac{1}{3+2\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{3+2\cos^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{\frac{3}{\cos^2\text{x}}+\frac{2\cos^2\text{x}}{\cos^2\text{x}}}$
$=\int\frac{\sec^2\text{x}}{2\sec^2\text{x}+2}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{3(1+\tan^2\text{x})+2}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{3+3\tan^2\text{x}+2}\text{ dx}$
$=\int\frac{\sec^2\text{x}}{5+3\tan^2\text{x}}\ \text{dx}$
Let $\sqrt{3}\tan\text{x}=\text{t}$
$\sqrt{3}\sec^2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\frac{1}{\sqrt{3}}\int\frac{\text{dt}}{(\sqrt{5})^2+\text{t}^2}$
$=\frac{1}{\sqrt{3}+\sqrt{5}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{5}}\Big)+\text{C}$
$\text{I}=\frac{1}{\sqrt{15}}\tan^{-1}\Big(\frac{\sqrt{3}\tan\text{x}}{\sqrt{5}}\Big)+\text{C}$
View full question & answer→Question 465 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2\tan^{-1}\text{x}}{1+\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\Big(\frac{\text{x}^2\tan^{-1}\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$=\int\Big(\frac{\text{x}^2+1-1}{\text{x}^2+1}\Big)\tan^{-1}\text{x dx}$
$=\int\Big(1-\frac{1}{\text{x}^2+1}\Big)\tan^{-1}\text{x dx}$
$=\int1.\tan^{-1}\text{x dx}-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
$=\Big[\tan^{-1}\text{x}\int1\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)\int1\text{dx}\Big\}\text{dx}\Big]-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
$=\Big[\tan^{-1}\text{x}\times\text{x}-\int\frac{\text{x}}{1+\text{x}^2}\text{dx}\Big]-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
Putting $\text{x}^2+1=\text{t}$ in the first integral and $\tan^{-1}\text{x}=\text{p}$ in the second integral
$\Rightarrow2\text{x dx = dt} $ and $\frac{1}{1+\text{x}^2}\text{dx = dp}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ and $\frac{1}{1+\text{x}^2}\text{dx = dp}$
$\therefore \text{I}=\tan^{-1}\text{x.x}-\frac{1}{2}\int\frac{\text{dt}}{\text{t}}-\int\text{p.dp}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\ln|\text{t}|-\frac{\text{P}^2}{2}+\text{C}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\ln|1+\text{x}^2|-\frac{(\tan^{-1}\text{x})^2}{2}+\text{C}$ $[\therefore \text{t}=\text{x}^2+1\text{ and}\text{ p}=\tan^{-1}\text{x}]$
View full question & answer→Question 475 Marks
Evaluate the following integrals:$\int(\text{x}+1)\text{e}^{\text{x}}\log(\text{xe}^{\text{x}})\text{dx}$
Answer$\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{xe}^{\text{x}})\text{dx}$ Let $\text{x e}^{\text{x}}=\text{t}$ $\Rightarrow\big(\text{x.e}^{\text{x}}+1.\text{e}^{\text{x}}\big)\text{dx=dt}$ $\therefore\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{x e}^\text{x})\text{dx}=\int1.\log (\text{t})\text{dt}$ $=\log\text{t}\int1\text{dt}-\int\big\{\frac{\text{d}}{\text{dt}}(\log\text{t)}-\int1\text{dt}\Big\}\text{dt}$ $=\log(\text{t})\times\text{t}-\int\frac{1}{\text{t}}\times\text{t dt}$ $=\text{t}\log(\text{t})-\text{t}+\text{C} \dots(1)$Substituting the value of t in eq (1)
$\Rightarrow\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{x e}^{\text{x}})\text{dx} = (\text{x e}^{\text{x}}).\log(\text{x e}^{\text{x}})-\text{x e}^{\text{x}}+\text{C}$ $=\text{x e}^{\text{x}}\Big\{\log(\text{x e}^{\text{x}})-1\Big\}+\text{C}$
View full question & answer→Question 485 Marks
Evaluate the following integrals:$\int\text{x}\tan^2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\tan^2\text{x dx}$
$=\int\text{x}(\sec^2\text{x}-1)\text{dx}$
$=\int\text{x}\sec^2\text{x dx}-\int\text{x dx}$
$=[\text{x}\int\sec^2\text{x dx}-\int(1\int\sec^2\text{x dx})\text{dx}]-\frac{\text{x}^2}{2}$
$=\text{x}\tan\text{x}-\int\tan\text{x dx}-\frac{\text{x}^2}{2}$
$\text{I}=\text{x}\tan\text{x}-\log\sec\text{x}-\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 495 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{7-3\text{x}-2\text{x}^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{7-3\text{x}-2\text{x}^2}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-\frac{3}{2}\text{x}-\text{x}^2}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-\big(\text{x}^2-\frac{3}{2}\text{x}\big)}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt2}\Big)^2-\Big(\text{x}^2+\frac{3}{2}\text{x}+\big(\frac{3}{4}\big)^2-\big(\frac{3}{4}\big)^2\Big)}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt2}\Big)^2-\big(\text{x}+\frac{3}{4}\big)^2+\frac{9}{16}}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}+\frac{9}{16}-\big(\text{x}+\frac{3}{4}\big)^2}}$ $=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{56+9}{16}-\big(\text{x}+\frac{3}{4}\big)^2}}$$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt{65}}{4}\Big)^2-\big(\text{x}+\frac{3}{4}\big)^2}}$
$=\frac{1}{\sqrt2}\sin^{-1}\Bigg[\frac{\text{x}+\frac{3}{4}}{\frac{\sqrt{65}}{4}}{}\Bigg]+\text{C}$ $=\frac{1}{\sqrt2}\sin^{-1}\Big[\frac{4\text{x}+3}{\sqrt{65}}\Big]+\text{C}$
View full question & answer→Question 505 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{3\text{x}^2+5\text{x}+7}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{3\text{x}^2+5\text{x}+7}}$
$=\int\frac{\text{dx}}{\sqrt{3\big(\text{x}^2+\frac{5}{3}\text{x}+\frac{7}{3}}\big)}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\text{x}^2+\frac{5}{3}\text{x}+\big(\frac{5}{6}\big)^2-\big(\frac{5}{6}\big)^2+\frac{7}{3}}}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\big(\text{x}+\frac{5}{6}\big)^2-\frac{25}{36}+\frac{7}{3}}}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\big(\text{x}+\frac{5}{6}\big)^2+\frac{-25+84}{36}}}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\big(\text{x}+\frac{5}{6}\big)^2+\frac{59}{36}}}$
$=\frac{1}{\sqrt3}\int\frac{\text{dx}}{\sqrt{\big(\text{x}+\frac{5}{6}\big)^2+\Big(\sqrt{\frac{59}{36}}\Big)^2}}$
$=\frac{1}{\sqrt3}\log\Bigg|\text{x}+\frac{5}{6}+\sqrt{\big(\text{x}+\frac{5}{6}\big)^2+\frac{59}{36}}\Bigg|+\text{C}$
$=\frac{1}{\sqrt3}\log\Bigg|\text{x}+\frac{5}{6}+\sqrt{\text{x}^2+\frac{5}{3}\text{x}+\frac{7}{3}}\Bigg|+\text{C}$
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