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Maths (commerce) Solve the Following Question.(3 Marks)

Integration (p-1) · Maharashtra Board STD 12 Commerce / Arts (MSBSHSE - English Medium). 18 questions.

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Solve the Following Question.(3 Marks)

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18 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Evalute : $\int \log \left(x^2+x\right) d x$
Answer
$
\begin{aligned}
& \text { Let } I=\int \log \left(x^2+x\right) d x=\int\left[\log \left(x^2+x\right)\right] \cdot 1 d x \\
= & {\left[\log \left(x^2+x\right)\right] \int 1 d x-\int\left[\frac{d}{d x}\left\{\log \left(x^2+x\right)\right\} \cdot \int d x\right] d x } \\
= & {\left[\log \left(x^2+x\right)\right] \cdot x-\int \frac{1}{x^2+x} \cdot \frac{d}{d x}\left(x^2+x\right) \times x d x } \\
= & x \log \left(x^2+x\right)-\int \frac{1}{x(x+1)} \cdot(2 x+1) \cdot x d x \\
= & x \log \left(x^2+x\right)-\int \frac{2 x+1}{x+1} d x \\
= & x \log \left(x^2+x\right)-\int \frac{2(x+1)-1}{x+1} d x \\
= & x \log \left(x^2+x\right)-\int\left(2-\frac{1}{x+1}\right) d x \\
= & x \log \left(x^2+x\right)-2 \int 1 d x+\int \frac{1}{x+1} d x \\
= & x \log \left(x^2+x\right)-2 x+\log |x+1|+c .
\end{aligned}
$
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Question 23 Marks
Evalute : $\int \frac{d x}{x\left[(\log x)^2+4 \log x-1\right]}$
Answer
Let $I=\int \frac{d x}{x\left[(\log x)^2+4 \log x-1\right]}$
$
=\int \frac{1}{(\log x)^2+4 \log x-1} \cdot \frac{1}{x} d x
$
Put $\log x=t \quad \therefore \frac{1}{x} d x=d t$
$
\begin{aligned}
\therefore I & =\int \frac{1}{t^2+4 t-1} d t \\
& =\int \frac{1}{\left(t^2+4 t+4\right)-5} d t \\
& =\int \frac{1}{(t+2)^2-(\sqrt{5})^2} d t \\
& =\frac{1}{2 \sqrt{5}} \log \left|\frac{t+2-\sqrt{5}}{t+2+\sqrt{5}}\right|+c
\end{aligned}
$
$
=\frac{1}{2 \sqrt{5}} \log \left|\frac{\log x+2-\sqrt{5}}{\log x+2+\sqrt{5}}\right|+c
$
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Question 33 Marks
Evalute : $\int \frac{2 e^x-3}{4 e^x+1} d x$
Answer
Let $I =\int \frac{2 e ^{ x }-3}{4 e ^{ x }+1} dx$
Let $2 e ^{ x }-3= A \left(4 e ^{ x }+1\right)+ B \frac{ d }{ dx }\left(4 e ^{ x }+1\right)$
$
\therefore 2 e ^{ x }-3=(4 A +4 B ) e ^{ x }+ A
$
Comparing the coefficients of $e ^{ x }$ and constant term on both sides, we get
$
4 A+4 B=2 \text { and } A=-3
$
Solving these equations, we get
$
\begin{aligned}
& B=\frac{7}{2} \\
& \therefore I=\frac{-3\left(4 e^x+1\right)+\frac{7}{2}\left(4 e^x\right)}{4 e^x+1} d x \\
& =-3 \int d x+\frac{7}{2} \int \frac{4 e^x}{4 e^x+1} d x \\
& \therefore I=-3 x+\frac{7}{2} \log \left|4 e^x+1\right|+c \quad \ldots\left[\because \int \frac{f \prime(x)}{f(x)} d x=\log |f(x)|+c\right]
\end{aligned}
$
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Question 43 Marks
Evalute : $\int \frac{1}{x\left(x^n+1\right)} d x$
Answer
Let $I=\int \frac{1}{x\left(x^n+1\right)} d x$
$
=\int \frac{x^{n-1}}{x^n\left(x^n+1\right)} d x
$
Put $x^n=t \quad \therefore n x^{n-1} d x=d t$
$
\begin{aligned}
& \therefore x^{n-1} d x=\frac{d t}{n} \\
& \begin{aligned}
\therefore I & =\int \frac{1}{t(t+1)} \cdot \frac{d t}{n} \\
& =\frac{1}{n} \int \frac{(t+1)-t}{t(t+1)} d t \\
& =\frac{1}{n} \int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t \\
& =\frac{1}{n}\left[\int \frac{1}{t} d t-\int \frac{1}{t+1} d t\right] \\
& =\frac{1}{n}[\log |t|-\log |t+1|]+c \\
& =\frac{1}{n} \log \left|\frac{t}{t+1}\right|+c=\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|+c .
\end{aligned}
\end{aligned}
$
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Question 53 Marks
Evalute : $\int \frac{1}{x\left(x^5+1\right)} d x$
Answer
$
\begin{aligned}
& \text { Let } I =\int \frac{1}{x\left(x^5+1\right)} d x \\
& =\int \frac{x^4}{x^5\left(x^5+1\right)} d x
\end{aligned}
$
Put $x^5=t$. Then $5 x^4 d x=d t$
$
\begin{array}{rl}
\therefore x^4 & d x=\frac{d t}{5} \\
\therefore I & =\int \frac{1}{t(t+1)} \cdot \frac{d t}{5} \\
& =\frac{1}{5} \int \frac{(t+1)-t}{t(t+1)} d t \\
& =\frac{1}{5} \int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t \\
& =\frac{1}{5}\left[\int \frac{1}{t} d t-\int \frac{1}{t+1} d t\right] \\
& =\frac{1}{5}[\log |t|-\log |t+1|]+c
\end{array}
$
$
=\frac{1}{5} \log \left|\frac{t}{t+1}\right|+c=\frac{1}{5} \log \left|\frac{x^5}{x^5+1}\right|+c \text {. }
$
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Question 63 Marks
Evalute : $\int \frac{2 x+1}{(x+1)(x-2)} d x$
Answer
Let $I =\int \frac{2 x+1}{(x+1)(x-2)} d x$
Let $\frac{2 x+1}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}$
$
\therefore 2 x +1= A ( x -2)+ B ( x +1)
$
Put $x+1=0$, i.e. $x=-1$, we get
$
\begin{aligned}
& 2(-1)+1= A (-3)+ B (0) \\
& \therefore A =\frac{1}{3}
\end{aligned}
$
Put $x-2=0$, i.e. $x=2$, we get
$
\begin{aligned}
& 2(2)+1= A (0)+ B (3) \\
& \therefore B =\frac{5}{3} \\
& \therefore \frac{2 x+1}{(x+1)(x-2)}=\frac{(1 / 3)}{x+1}+\frac{(5 / 3)}{x-2} \\
& \therefore I=\int\left[\frac{(1 / 3)}{x+1}+\frac{(5 / 3)}{x-2}\right] d x
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{3} \int \frac{1}{x+1} d x+\frac{5}{3} \int \frac{1}{x-2} d x \\
& =\frac{1}{3} \log |x+1|+\frac{5}{3} \log |x-2|+c .
\end{aligned}
$
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Question 73 Marks
Evalute : $\int x^2 e^{3 x} d x$
Answer
$
\begin{aligned}
& \text { Let } I=\int x^2 e^{3 x} d x \\
& =x^2 \int e^{3 x} d x-\int\left[\frac{d}{d x}\left(x^2\right) \int e^{3 x} d x\right] d x \\
& =x^2 \cdot\left(\frac{e^{3 x}}{3}\right)-\int 2 x \cdot \frac{e^{3 x}}{3} d x \\
& =\frac{x^2}{3} e^{3 x}-\frac{2}{3} \int x \cdot e^{3 x} d x \\
& =\frac{x^2}{3} e^{3 x}-\frac{2}{3}\left[x \int e^{3 x} d x-\int\left(\frac{d}{d x}(x) \int e^{3 x} d x\right) d x\right] \\
& =\frac{x^2 \cdot e^{3 x}}{3}-\frac{2}{3}\left[x \cdot \frac{e^{3 x}}{3}-\int 1 \cdot \frac{e^{3 x}}{3} d x\right] \\
& =\frac{x^2 \cdot e^{3 x}}{3}-\frac{2}{3}\left[\frac{1}{3} x e^{3 x}-\frac{1}{3} \int e^{3 x} d x\right] \\
& =\frac{x^2 \cdot e^{3 x}}{3}-\frac{2}{3}\left[\frac{1}{3} x e^{3 x}-\frac{1}{3} \cdot \frac{e^{3 x}}{3}\right]+c\\
& \therefore I=\frac{1}{3} x^2 \cdot e^{3 x}-\frac{2}{9} x e^{3 x}+\frac{2}{27} e^{3 x}+c \\
\end{aligned}
$
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Question 83 Marks
Evalute : $\int x^2 e^{4 x} d x$
Answer
$
\begin{aligned}
& \int x^2 e^{4 x} d x=x^2 \int e^{4 x} d x-\int\left[\frac{d}{d x}\left(x^2\right) \int e^{4 x} d x\right] d x \\
& =x^2 \cdot \frac{e^{4 x}}{4}-\int 2 x \cdot \frac{e^{4 x}}{4} d x \\
& =\frac{1}{4} x^2 e^{4 x}-\frac{1}{2} \int x e^{4 x} d x \\
& =\frac{1}{4} x^2 e^{4 x}-\frac{1}{2}\left[x \int e^{4 x} d x-\int\left\{\frac{d}{d x}(x) \int e^{4 x} d x\right\} d x\right] \\
& =\frac{1}{4} x^2 e^{4 x}-\frac{1}{2}\left[x \cdot \frac{e^{4 x}}{4}-\int 1 \cdot \frac{e^{4 x}}{4} d x\right] \\
& =\frac{1}{4} x^2 e^{4 x}-\frac{1}{8} x \cdot e^{4 x}+\frac{1}{8} \int e^{4 x} d x \\
& =\frac{1}{4} x^2 e^{4 x}-\frac{1}{8} x e^{4 x}+\frac{1}{8} \cdot \frac{e^{4 x}}{4}+c \\
& =\frac{1}{4} e^{4 x}\left[x^2-\frac{x}{2}+\frac{1}{8}\right]+c
\end{aligned}
$
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Question 93 Marks
Evalute : $\int \frac{\log x}{(1+\log x)^2} d x$
Answer
Let $I=\int \frac{\log x}{(1+\log x)^2} d x$
Put $\log x=t \quad \therefore x=e^t$
$
\therefore d x=e^t d t
$
$
\begin{aligned}
\therefore I & =\int \frac{t}{(1+t)^2} \cdot e^t d t \\
& =\int e^t\left[\frac{(1+t)-1}{(1+t)^2}\right] d t \\
& =\int e^t\left[\frac{1}{1+t}-\frac{1}{(1+t)^2}\right] d t
\end{aligned}
$
Let $f(t)=\frac{1}{1+t}$.
$
\begin{aligned}
\therefore f^{\prime}(t) & =\frac{d}{d t}(1+t)^{-1}=-1(1+t)^{-2}(0+1) \\
& =\frac{-1}{(1+t)^2}
\end{aligned}
$
$
\begin{aligned}
\therefore I & =\int e^t\left[f(t)+f^{\prime}(t)\right] d t \\
& =e^t \cdot f(t)+c=e^t \times \frac{1}{1+t}+c=\frac{x}{1+\log x}+c .
\end{aligned}
$
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Question 103 Marks
Evalute : $\int \frac{x^3}{16 x^8-25} d x$
Answer
Let $I=\int \frac{x^3}{16 x^8-25} d x$
Put $x^4=t \quad \therefore 4 x^3 d x=d t$
$
\therefore x^3 d x=\frac{d t}{4}
$
$
\begin{aligned}
\therefore I & =\int \frac{1}{16 t^2-25} \cdot \frac{d t}{4} \\
& =\frac{1}{4} \times \frac{1}{16} \int \frac{1}{t^2-\frac{25}{16}} d t \\
& =\frac{1}{64} \int \frac{1}{t^2-\left(\frac{5}{4}\right)^2} d t
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{64} \times \frac{1}{2 \times \frac{5}{4}} \log \left|\frac{t-\frac{5}{4}}{t+\frac{5}{4}}\right|+c \\
& =\frac{1}{160} \log \left|\frac{4 t-5}{4 t+5}\right|+c \\
& =\frac{1}{160} \log \left|\frac{4 x^4-5}{4 x^4+5}\right|+c .
\end{aligned}
$
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Question 113 Marks
Evalute : $\int \frac{x}{4 x^4-20 x^2-3} d x$
Answer
Let $I=\int \frac{x}{4 x^4-20 x^2-3} d x$
Put $x^2=t \quad \therefore 2 x d x=d t$
$\therefore x d x=\frac{d t}{2}$
$
\therefore I=\int \frac{1}{4 t^2-20 t-3} \cdot \frac{d t}{2}
$
$=\frac{1}{2} \times \frac{1}{4} \int \frac{1}{t^2-5 t-\frac{3}{4}} d t$
$=\frac{1}{8} \int \frac{1}{\left(t^2-5 t+\frac{25}{4}\right)-\frac{25}{4}-\frac{3}{4}} d t$
$=\frac{1}{8} \int \frac{1}{\left(t-\frac{5}{2}\right)^2-(\sqrt{7})^2} d t$
$=\frac{1}{8} \times \frac{1}{2 \sqrt{7}} \log \left|\frac{t-\frac{5}{2}-\sqrt{7}}{t-\frac{5}{2}+\sqrt{7}}\right|+c$
$=\frac{1}{16 \sqrt{7}} \log \left|\frac{2 t-5-2 \sqrt{7}}{2 t-5+2 \sqrt{7}}\right|+c$
$=\frac{1}{16 \sqrt{7}} \log \left|\frac{2 x^2-5-2 \sqrt{7}}{2 x^2-5+2 \sqrt{7}}\right|+c$.
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Question 123 Marks
Evalute : $\int \frac{1}{4 x^2-20 x+17} d x$
Answer
$
\begin{aligned}
& \int \frac{1}{4 x^2-20 x+17} d x \\
= & \frac{1}{4} \int \frac{1}{x^2-5 x+\frac{17}{4}} d x \\
= & \frac{1}{4} \int \frac{1}{\left(x^2-5 x+\frac{25}{4}\right)-\frac{25}{4}+\frac{17}{4}} d x \\
& =\frac{1}{4} \int \frac{1}{\left(x-\frac{5}{2}\right)^2-(\sqrt{2})^2} d x
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{4} \times \frac{1}{2 \sqrt{2}} \log \left|\frac{x-\frac{5}{2}-\sqrt{2}}{x-\frac{5}{2}+\sqrt{2}}\right|+c \\
& =\frac{1}{8 \sqrt{2}} \log \left|\frac{2 x-5-2 \sqrt{2}}{2 x-5+2 \sqrt{2}}\right|+c .
\end{aligned}
$
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Question 133 Marks
Evalute : $\int \frac{2 e^x+5}{2 e^x+1} d x$
Answer
Let $I =\int \frac{2 e ^{ x }+5}{2 e ^{ x }+1} dx$
Let $2 e^x+5=A\left(2 e^x+1\right)+B \frac{d}{d x}\left(2 e^x+1\right)$
$
\begin{aligned}
& =2 A e^x+A+B\left(2 e^x\right) \\
& \therefore 2 e^x+5=(2 A+2 B) e^x+A
\end{aligned}
$
Comparing the coefficients of $e ^{ x }$ and constant term on both sides, we get
$
2 A +2 B =2 \text { and } A =5
$
Solving these equations, we get
$
\begin{aligned}
& B =-4 \\
& \therefore I =\int \frac{5\left(2 e ^{ x }+1\right)-4\left(2 e ^{ x }\right)}{2 e ^{ x }+1} dx \\
& =5 \int dx -4 \int \frac{2 e ^{ x }}{2 e ^{ x }+1} dx \\
& \therefore I =5 x -4 \log \left|2 e ^{ x }+1\right|+c \quad \ldots . .\left[\int \frac{ f \prime ( x )}{ f ( x )} dx =\log |f( x )|+ c \right]
\end{aligned}
$
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Question 143 Marks
Evalute : $\int \frac{20-12 e^x}{3 e^x-4} d x$
Answer
Let $I =\int \frac{20-12 e^x}{3 e^x-4} d x$
Put, Numerator $= A$ (Denominator $)+ B \left[\frac{d}{d x}\right.$ (Denominator) $]$
$
\begin{aligned}
& \therefore 20-12 e ^{ x }= A \left(3 e ^{ x }-4\right)+ B \left[\frac{d}{d x}\left(3 e ^{ x }-4\right)\right] \\
& \therefore 20-12 e ^{ x }= A \left(3 e ^{ x }-4\right)+ B \left(3 e ^{ x }-0\right) \\
& \therefore 20-12 e ^{ x }=(3 A +3 B ) e ^{ x }-4 A
\end{aligned}
$
Equating the coefficient of ex and constant on both sides, we get
$
\begin{aligned}
& 3 A+3 B=-12 \\
& \text { and }-4 A=20 \\
& \therefore A=-5
\end{aligned}
$
from (1), $3(-5)+3 B=-12$
$
\begin{aligned}
& \therefore 3 B =3 \\
& \therefore B =1 \\
& \therefore 20-12 e ^{ x }=-5\left(3 e ^{ x }-4\right)+\left(3 e ^{ x }\right) \\
& \therefore I=\int\left[\frac{-5\left(3 e^x-4\right)+\left(3 e^x\right)}{3 e^x-4}\right] d x \\
& \quad=\int\left(-5+\frac{3 e^x}{3 e^x-4}\right) d x \\
& \quad=-5 \int 1 d x+\int \frac{3 e^x}{3 e^x-4} d x \\
& \quad=-5 x+\log \left|3 e^x-4\right|+c \\
& \quad \ldots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]
\end{aligned}
$
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Question 153 Marks
Evalute : $\int \frac{x^5}{x^2+1} d x$
Answer
Let $I=\int \frac{x^5}{x^2+1} d x=\int \frac{x^4}{x^2+1} \cdot x d x$
$
=\int \frac{\left(x^2\right)^2}{x^2+1} \cdot x d x
$
Put $x^2+1=t \quad \therefore 2 x d x=d t$
$
\begin{aligned}
& \therefore x d x=\frac{d t}{2} \text { and } x^2=t-1 \\
& \begin{aligned}
\therefore I & =\int \frac{(t-1)^2}{t} \cdot \frac{d t}{2}=\frac{1}{2} \int\left(\frac{t^2-2 t+1}{t}\right) d t \\
& =\frac{1}{2} \int\left(t-2+\frac{1}{t}\right) d t \\
& =\frac{1}{2} \int t d t-\int 1 d t+\frac{1}{2} \int \frac{1}{t} d t \\
& =\frac{1}{2} \cdot \frac{t^2}{2}-t+\frac{1}{2} \log |t|+c
\end{aligned}
\end{aligned}
$
$
=\frac{1}{4}\left(x^2+1\right)^2-\left(x^2+1\right)+\frac{1}{2} \log \left|x^2+1\right|+c .$
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Question 163 Marks
Evalute : $\int \frac{1}{x\left(x^6+1\right)} d x$
Answer
Let $I=\int \frac{1}{x\left(x^6+1\right)} d x$
$
=\int \frac{x^5}{x^6\left(x^6+1\right)} d x
$
Put $x^6=t \quad \therefore 6 x^5 d x=d t$
$
\therefore x^5 d x=\frac{1}{6} d t
$
$
\begin{aligned}
\therefore I & =\int \frac{1}{t(t+1)} \cdot \frac{d t}{6} \\
& =\frac{1}{6} \int \frac{(t+1)-t}{t(t+1)} d t=\frac{1}{6} \int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t \\
& =\frac{1}{6}\left[\int \frac{1}{t} d t-\int \frac{1}{t+1} d t\right] \\
& =\frac{1}{6}[\log (t)-\log |t+1|]+c \\
& =\frac{1}{6} \log \left|\frac{t}{t+1}\right|+c=\frac{1}{6} \log \left|\frac{x^6}{x^6+1}\right|+c .
\end{aligned}
$
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Question 173 Marks
If $f(x)=\frac{x^2}{2}-k x+1, f(0)=2$ and $f(3)=5$, find $f(x)$.
Answer
By the definition of integral
$
\begin{aligned}
f(x) & =\int f^{\prime}(x) d x=\int\left(\frac{x^2}{2}-k x+1\right) d x \\
& =\frac{1}{2} \int x^2 d x-k \int x d x+\int 1 d x \\
& =\frac{1}{2}\left(\frac{x^3}{3}\right)-k\left(\frac{x^2}{2}\right)+x+c \\
\therefore f(x) & =\frac{x^3}{6}-\frac{k x^2}{2}+x+c
\end{aligned}
$
Now, $f(0)=2$ gives
$
\begin{aligned}
& f(0)=0-0+0+c=2 \quad \therefore c=2 \\
& \therefore \text { from (1), } f(x)=\frac{x^3}{6}-\frac{k x^2}{2}+x+2
\end{aligned}
$
Further $f(3)=5$ gives
$
\begin{aligned}
& f(3)=\frac{27}{6}-\frac{9 k}{2}+3+2=5 \\
& \therefore \frac{9 k}{2}=\frac{9}{2} \quad \therefore k=1 \\
& \therefore \text { from }(2), f(x)=\frac{x^3}{6}-\frac{x^2}{2}+x+2 .
\end{aligned}
$
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Question 183 Marks
If $f(x)=4 x^3-3 x^2+2 x+k, f(0)=-1$ and $f(1)=4$, find $f(x)$.
Answer
By the definition of integral
$
\begin{aligned}
& f(x)=\int f^{\prime}(x) d x \\
& =\int\left(4 x^3-3 x^2+2 x+k\right) d x \\
& =4 \int x^3 d x-3 \int x^2 d x+2 \int x d x+k \int 1 d x \\
& =4\left(\frac{x^4}{4}\right)-3\left(\frac{x^3}{3}\right)+2\left(\frac{x^2}{2}\right)+k x+c \\
& \therefore f(x)=x^4-x^3+x^2+k x+c
\end{aligned}
$
Now, $f (0)=1$ gives
$
\begin{aligned}
& f(0)=0-0+0+0+c=1 \\
& \therefore c=1
\end{aligned}
$
$\therefore$ from (1), $f ( x )= x ^4- x ^3+ x ^2+ kx +1$
Further $f(1)=4$ gives
$
\begin{aligned}
& f(1)=1-1+1+k+1=4 \\
& \therefore k=2 \\
& \therefore \text { from }(2), f(x)=x^4-x^3+x^2+2 x+1 .
\end{aligned}
$
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