Solve the Following Question.(5 Marks)

Question 15 Marks

Evalute : $\int \frac{1+\log x}{x(3+\log x)(2+3 \log x)} d x$

Answer

Let $I=\int \frac{1+\log x}{x(3+\log x)(2+3 \log x)} d x$
$
=\int \frac{1+\log x}{(3+\log x)(2+3 \log x)} \cdot \frac{1}{x} d x
$
Put $\log x=t \quad \therefore \frac{1}{x} d x=d t$
$
\therefore I=\int \frac{1+t}{(3+t)(2+3 t)} d t
$
Let $\frac{1+t}{(3+t)(2+3 t)}=\frac{A}{3+t}+\frac{B}{2+3 t}$
$
\therefore 1+t=A(2+3 t)+B(3+t)
$
Put $3+t=0$, i.e. $t=-3$, we get
$
1-3=A(-7)+B(0)
$
$
\therefore-2=-7 A \quad \therefore A=\frac{2}{7}
$
Put $2+3 t=0$, i.e. $t=-\frac{2}{3}$, we get
$
1-\frac{2}{3}=A(0)+B\left(\frac{7}{3}\right)
$
$
\therefore \frac{1}{3}=\frac{7}{3} B \quad \therefore B=\frac{1}{7}
$
$
\therefore \frac{1+t}{(3+t)(2+3 t)}=\frac{\left(\frac{2}{7}\right)}{3+t}+\frac{\left(\frac{1}{7}\right)}{2+3 t}
$
$\begin{aligned} \therefore I & =\int\left[\frac{\left(\frac{2}{7}\right)}{3+t}+\frac{\left(\frac{1}{7}\right)}{2+3 t}\right] d t \\ & =\frac{2}{7} \int \frac{1}{3+t} d t+\frac{1}{7} \int \frac{1}{2+3 t} d t \\ & =\frac{2}{7} \log |3+t|+\frac{1}{7} \cdot \frac{\log |2+3 t|}{3}+c \\ & =\frac{2}{7} \log |3+\log x|+\frac{1}{21} \log |2+3 \log x|+c .\end{aligned}$

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Question 25 Marks

Evalute : $\int \frac{2 x^3-3 x^2-9 x+1}{2 x^2-x-10} d x$

Answer

$
\begin{aligned}
& \text { Let } I=\int \frac{2 x^3-3 x^2-9 x+1}{2 x^2-x-10} d x \\
&\left.2 x^2-x-10\right) 2 x^3-3 x^2-9 x+1(x-1 \\
& \frac{2 x^3-x^2-10 x}{+}+ \\
&-2 x^2+x+1 \\
&+2 x^2+x+10 \\
&+-9
\end{aligned}
$
$
\begin{array}{rl}
\therefore I & 2 x^3-3 x^2-9 x+1=(x-1)\left(2 x^2-x-10\right)-9 \\
\therefore I & =\int\left[\frac{(x-1)\left(2 x^2-x-10\right)-9}{2 x^2-x-10}\right] d x \\
& =\int\left[(x-1)-\frac{9}{2 x^2-x-10}\right] d x \\
& =\int(x-1) d x-\frac{9}{2} \int \frac{1}{x^2-\frac{1}{2} x-5} d x \\
& =\int x d x-\int 1 d x-\frac{9}{2} \int \frac{1}{\left(x^2-\frac{1}{2} x+\frac{1}{16}\right)^{-}-\frac{1}{16}-5} d x \\
& =\int x d x-\int 1 d x-\frac{9}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2-\left(\frac{9}{4}\right)^2} d x
\end{array}
$
$
=\frac{x^2}{2}-x-\frac{9}{2} \times \frac{1}{2 \times \frac{9}{4}} \log \left|\frac{x-\frac{1}{4}-\frac{9}{4}}{x-\frac{1}{4}+\frac{9}{4}}\right|+c_1
$
$\begin{aligned} & =\frac{x^2}{2}-x-\log \left|\frac{x-\frac{5}{2}}{x+2}\right|+c_1 \\ & =\frac{x^2}{2}-x-\log \left|\frac{2 x-5}{2(x+2)}\right|+c_1 \\ & =\frac{x^2}{2}-x+\log \left|\frac{2(x+2)}{2 x-5}\right|+c_1 \\ & =\frac{x^2}{2}-x+\log \left|\frac{x+2}{2 x-5}\right|+\log 2+c_1 \\ & =\frac{x^2}{2}-x+\log \left|\frac{x+2}{2 x-5}\right|+c, \text { where } c_1=\log 2+c_1\end{aligned}$

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Question 35 Marks

Evalute : $\int \frac{3 x-1}{2 x^2-x-1} d x$

Answer

$
\text { Let } \begin{aligned}
I & =\int \frac{3 x-1}{2 x^2-x-1} d x \\
& =\int \frac{3 x-1}{(x-1)(2 x+1)} d x
\end{aligned}
$
Let $\frac{3 x-1}{(x-1)(2 x+1)}=\frac{A}{x-1}+\frac{B}{2 x+1}$
$
\therefore 3 x-1=A(2 x+1)+B(x-1)
$
Put $x-1=0$, i.e. $x=1$, we get
$
\begin{aligned}
& 3(1)-1=A(3)+B(0) \quad \therefore 2=3 A \quad \therefore A=\frac{2}{3} \\
\end{aligned}
$
Put $2 x+1=0$, i.e. $x=-\frac{1}{2}$, we get
$
\begin{aligned}
& 3\left(-\frac{1}{2}\right)-1=A(0)+B\left(-\frac{3}{2}\right) \\
& \therefore-\frac{5}{2}=-\frac{3}{2} B \quad \therefore B=\frac{5}{3} \\
& \therefore \frac{3 x-1}{(x-1)(2 x+1)}=\frac{\left(\frac{2}{3}\right)}{x-1}+\frac{\left(\frac{5}{3}\right)}{2 x+1} \\
& \therefore I=\int\left[\frac{\left(\frac{2}{3}\right)}{x-1}+\frac{\left(\frac{5}{3}\right)}{2 x+1}\right] d x \\
& \quad=\frac{2}{3} \int \frac{1}{x-1} d x+\frac{5}{3} \int \frac{1}{2 x+1} d x \\
& \quad=\frac{2}{3} \log |x-1|+\frac{5}{3} \cdot \frac{\log |2 x+1|}{2}+c \\
& \quad=\frac{2}{3} \log |x-1|+\frac{5}{6} \log |2 x+1|+c .
\end{aligned}
$

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Question 45 Marks

Evalute : $\int \frac{5 x^2+20 x+6}{x^3+2 x^2+x} d x$

Answer

Let $I=\int \frac{5 x^2+20 x+6}{x^3+2 x^2+x} d x$
$
\begin{aligned}
& =\int \frac{5 x^2+20 x+6}{x\left(x^2+2 x+1\right)} d x \\
& =\int \frac{5 x^2+20 x+6}{x(x+1)^2} d x
\end{aligned}
$
Let $\frac{5 x^2+20 x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$
$
\therefore 5 x ^2+20 x +6= A ( x +1)^2+ Bx ( x +1)+ Cx
$
Put $x=0$, we get
$
\begin{aligned}
& 0+0+6= A (1)+ B (0)(1)+ C (0) \\
& \therefore A =6
\end{aligned}
$
Put $x+1=0$, i.e. $x=-1$, we get
$
\begin{aligned}
& 5(1)+20(-1)+6= A (0)+ B (-1)(0)+ C (-1) \\
& \therefore-9=- C \\
& \therefore C =9
\end{aligned}
$
Put $x =1$, we get
$
\begin{aligned}
& 5(1)+20(1)+6=A(4)+B(1)(2)+C(1) \\
& \text { But } A=6 \text { and } C=9 \\
& \therefore 31=24+2 B+9 \\
& \therefore B=-1 \\
& \therefore \frac{5 x^2+20 x+6}{x(x+1)^2}=\frac{6}{x}-\frac{1}{x+1}+\frac{9}{(x+1)^2} \\
& \therefore I=\int\left[\frac{6}{x}-\frac{1}{x+1}+\frac{9}{(x+1)^2}\right] d x
\end{aligned}
$
$
\begin{aligned}
& =6 \int \frac{1}{x} d x-\int \frac{1}{x+1} d x+9 \int(x+1)^{-2} d x \\
& =6 \log |x|-\log |x+1|+9 \cdot \frac{(x+1)^{-1}}{-1}+c
\end{aligned}
$
$
=6 \log |x|-\log |x+1|-\frac{9}{x+1}+c .
$

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Question 55 Marks

Evalute : $\int \frac{3 x-2}{(x+1)^2(x+3)} d x$

Answer

Let $I =\int \frac{3 x-2}{(x+1)^2(x+3)} d x$
Let $\frac{3 x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}$
$
\therefore 3 x -2= A ( x +1)( x +3)+ B ( x +3)+ C ( x +1)^2
$
Put $x+1=0$, i.e. $x=-1$, we get
$
3(-1)-2= A (0)(2)+ B (2)+ C (0)
$
$
\therefore-5=2 B
$
$
\therefore B =-\frac{5}{2}
$
Put $x+3=0$, i.e. $x=-3$, we get
$
\begin{aligned}
& 3(-3)-2= A (-2)(0)+ B (0)+ C (4) \\
& \therefore-11=4 C \\
& \therefore C =-\frac{11}{4}
\end{aligned}
$
Put $x =0$, we get
$
\begin{aligned}
& 3(0)-2= A (1)(3)+ B (3)+ C (1) \\
& \therefore-2=3 A +3 B + C
\end{aligned}
$
But $B =-\frac{5}{2}$ and $C =-\frac{11}{4}$
$
\begin{aligned}
& \therefore-2=3 A+3\left(-\frac{5}{2}\right)-\frac{11}{4} \\
& \therefore 3 A=-2+\frac{15}{2}+\frac{11}{4}=\frac{-8+30+11}{4}=\frac{33}{4} \\
& \therefore A=\frac{11}{4}\\
& \therefore \frac{3 x-2}{(x+1)^2(x+3)}=\frac{\left(\frac{11}{4}\right)}{x+1}+\frac{\left(-\frac{5}{2}\right)}{(x+1)^2}+\frac{\left(-\frac{11}{4}\right)}{(x+3)}
\end{aligned}
$
$\begin{aligned} \therefore I & =\int\left[\frac{\left(\frac{11}{4}\right)}{x+1}+\frac{\left(-\frac{5}{2}\right)}{(x+1)^2}+\frac{\left(-\frac{11}{4}\right)}{x+3}\right] d x \\ & =\frac{11}{4} \int \frac{1}{x+1} d x-\frac{5}{2} \int(x+1)^{-2} d x-\frac{11}{4} \int \frac{1}{x+3} d x \\ & =\frac{11}{4} \log |x+1|-\frac{5}{2} \cdot \frac{(x+1)^{-1}}{-1}-\frac{11}{4} \log |x+3|+c \\ & =\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+c .\end{aligned}$

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Question 65 Marks

Evalute : $\int \frac{x}{(x-1)^2(x+2)} d x$

Answer

Let $I =\int \frac{x}{(x-1)^2(x+2)} d x$
Let $\frac{x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}$
$
\therefore x = A ( x -1)( x +2)+ B ( x +2)+ C ( x -1)^2
$
Put $x-1=0$, i.e. $x=1$, we get
$
\begin{aligned}
& 1= A (0)(3)+ B (3)+ C (0) \\
& \therefore B =\frac{1}{3}
\end{aligned}
$
Put $x+2=0$, i.e. $x=-2$, we get
$
\begin{aligned}
& -2= A (-3)(0)+ B (0)+ C (9) \\
& \therefore C =-\frac{2}{9}
\end{aligned}
$
Put $x =-1$, we get,
$
-1=A(-2)(1)+B(1)+C(4)
$
But $B =\frac{1}{3}$ and $C =-\frac{2}{9}$
$
\begin{aligned}
& \therefore-1=-2 A+\frac{1}{3}-\frac{8}{9} \\
& \therefore 2 A=-\frac{5}{9}+1=\frac{4}{9} \quad \therefore A=\frac{2}{9}
\end{aligned}
$
$
\begin{aligned}
& \therefore \frac{x}{(x-1)^2(x+2)}=\frac{(2 / 9)}{x-1}+\frac{(1 / 3)}{(x-1)^2}+\frac{(-2 / 9)}{x+2} \\
& \therefore I=\int\left[\frac{(2 / 9)}{x-1}+\frac{(1 / 3)}{(x-1)^2}+\frac{(-2 / 9)}{x+2}\right] d x \\
& \quad=\frac{2}{9} \int \frac{1}{x-1} d x+\frac{1}{3} \int(x-1)^{-2} d x-\frac{2}{9} \int \frac{1}{x+2} d x
\end{aligned}
$
$
=\frac{2}{9} \log |x-1|+\frac{1}{3} \cdot \frac{(x-1)^{-1}}{-1}-\frac{2}{9} \log |x+2|+c
$
$
=\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+c \text {. }
$

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Question 75 Marks

Evalute : $\int \frac{2 x+1}{x(x-1)(x-4)} d x$

Answer

Let $I =\int \frac{2 x+1}{x(x-1)(x-4)} d x$
Let $\int \frac{2 x+1}{x(x-1)(x-4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-4}$
$
\therefore 2 x +1= A ( x -1)( x -4)+ Bx ( x -4)+ Cx ( x -1)
$
Put $x =0$, we get
$
\begin{aligned}
& 2(0)+1= A (-1)(-4)+ B (0)(-4)+ C (0)(-1) \\
& \therefore 1=4 A \\
& \therefore A =\frac{1}{4}
\end{aligned}
$
Put $x-1=0$, i.e. $x=1$, we get
$
\begin{aligned}
& 2(1)+1= A (0)(-3)+ B (1)(-3)+ C (1)(0) \\
& \therefore 3=-3 B \\
& \therefore B =-1
\end{aligned}
$
Put $x-4=0$, i.e $x=4$, we get
$
\begin{aligned}
& 2(4)+1= A (3)(0)+ B (4)(0)+ C (4)(3) \\
& \therefore 9=12 C \\
& \therefore C =\frac{3}{4}
\end{aligned}
$
$
\therefore \frac{2 x+1}{x(x-1)(x-4)}=\frac{\left(\frac{1}{4}\right)}{x}+\frac{(-1)}{x-1}+\frac{\left(\frac{3}{4}\right)}{x-4}
$
$
\begin{aligned}
\therefore I & =\int\left[\frac{\left(\frac{1}{4}\right)}{x}+\frac{(-1)}{x-1}+\frac{\left(\frac{3}{4}\right)}{x-4}\right] d x \\
& =\frac{1}{4} \int \frac{1}{x} d x-\int \frac{1}{x-1} d x+\frac{3}{4} \int \frac{1}{x-4} d x \\
& =\frac{1}{4} \log |x|-\log |x-1|+\frac{3}{4} \log |x-4|+c .
\end{aligned}
$

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Question 85 Marks

Evalute : $\int \frac{3 e^x+4}{2 e^x-8} d x$

Answer

Let $I =\int \frac{3 e^x+4}{2 e^x-8} d x$
Put, Numerator $= A$ (Denominator $)+ B \left[\frac{d}{d x}\right.$ (Denominator $\left.)\right]$
$
\begin{aligned}
& \therefore 3 e ^{ x }+4= A \left(2 e ^{ x }-8\right)+ B \left[\frac{d}{d x}\left(2 e ^{ x }-8\right)\right] \\
& \therefore 3 e ^{ x }+4= A \left(2 e ^{ x }-8\right)+ B \left(2 e ^{ x }-0\right) \\
& \therefore 3 e ^{ x }+4=(2 A +2 B ) e ^{ x }-8 A
\end{aligned}
$
Equating the coefficient of ex and constant on both sides, we get
$
\begin{aligned}
& 2 A +2 B =3 \\
& \text { and }-8 A =4
\end{aligned}
$
$
\therefore A =-\frac{1}{2}
$
$\therefore$ from $(1), 2\left(-\frac{1}{2}\right)+2 B=3$
$
\therefore 2 B =4
$
$
\therefore B =2
$
$
\therefore 3 e^x+4=-\frac{1}{2}\left(2 e^x-8\right)+2\left(2 e^x\right)
$
$\therefore I=\int\left[\frac{-\frac{1}{2}\left(2 e^x-8\right)+2\left(2 e^x\right)}{2 e^x-8}\right] d x$
$
=\int\left[-\frac{1}{2}+\frac{2\left(2 e^x\right)}{2 e^x-8}\right] d x
$
$
=-\frac{1}{2} \int 1 d x+2 \int \frac{2 e^x}{2 e^x-8} d x
$
$
=-\frac{1}{2} x+2 \log \left|2 e^x-8\right|+c
$
$
\ldots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]
$

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Question 95 Marks

Evalute : $\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t$

Answer

Let $I =\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t$
Put, Numerator $= A ($ Denominator $)+ B \left[\frac{d}{d x}\right.$ (Denominator $\left.)\right]$
$
\begin{aligned}
& \therefore 3 e ^{2 t }+5= A \left(4 e ^{2 t }-5\right)+ B \left[\frac{d}{d t}\left(4 e ^{2 t }-5\right)\right] \\
& \therefore 3 e ^{2 t }+5= A \left(4 e ^{2 t }-5\right)+ B \left[4 e ^{2 t } \times 2-0\right] \\
& \therefore 3 e ^{2 t }+5=(4 A +8 B ) e ^{2 t }-5 A
\end{aligned}
$
Equating the coefficient of $e ^{2 t }$ and constant on both sides, we get
$
\begin{aligned}
& 4 A +8 B =3 \\
& \text { and }-5 A =5 \\
& \therefore A =-1 \\
& \therefore \text { from }(1), 4(-1)+8 B =3 \\
& \therefore 8 B =7 \\
& \therefore B =\frac{7}{8} \\
& \therefore 3 e^{2 t}+5=-\left(4 e^{2 t}-5\right)+\frac{7}{8}\left(8 e^{2 t}\right) \\
& \left.\therefore I=\int \frac{-\left(4 e^{2 t}-5\right)+\frac{7}{8}\left(8 e^{2 t}\right)}{4 e^{2 t}-5}\right] d t \\
& \quad=\int\left[-1+\frac{\frac{7}{8}\left(8 e^{2 t}\right)}{4 e^{2 t}-5}\right] d t \\
& \quad=-\int 1 d t+\frac{7}{8} \int \frac{8 e^{2 t}}{4 e^{2 t}-5} d t \\
& \quad=-t+\frac{7}{8} \log \left|4 e^{2 t}-5\right|+c \cdot \\
& ......\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right] .
\end{aligned}
$

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Maths (commerce) Solve the Following Question.(5 Marks)

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