Question

Match the following integrals in column I with their corresponding values in column II and choose the correct option from the codes given below.

Column I		Column II
A	$\int e^{2 x+3} d x$	1	$e^{\tan ^{-1} x}+C$
B	$\int \frac{x}{e^{x^2}} d x$	2	$\log \left|e^x+e^{-x}\right|+C$
C	$\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x$	3	$\frac{1}{2} e^{(2 x+3)}+C$
D	$\int \frac{e^{2 x}-1}{e^{2 x}+1} d x$	4	$-\frac{1}{2} e^{-\left(x^2\right)}+C$

• A. A - (1), B - (2), C - (3), D - (4)
• B. A - (3), B - (4), C - (1), D - (2)
• C. A - (1), B - (3), C - (2), D - (4)
• D. A - (3), B - (1), C - (2), D - (4)

Answer 1

Answer: (B)

A. $\int e^{(2 x+3)} d x$
Let $2 x+3=t \Rightarrow 2=\frac{d t}{d x} \Rightarrow d x=\frac{d t}{2}$
$\therefore \int e^{(2 x+3)} d x=\int e^t \frac{d t}{2}=\frac{1}{2} \int e^t d t$
$=\frac{1}{2} e^t+C=\frac{1}{2} e^{(2 x+3)}+C$
B. $\int \frac{x}{e^{x^2}} d x$
Let $x^2=t \Rightarrow 2 x=\frac{d t}{d x} \Rightarrow d x=\frac{d t}{2 x}$
$\therefore \int \frac{x}{e^{x^2}} d x=\int \frac{x}{e^t} \frac{d t}{2 x}=\frac{1}{2} \int e^{-t} d t=-\frac{1}{2} e^{-t}+C$
$=-\frac{1}{2} e^{-\left(x^2\right)}+C$
C. $\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x$
Let $\tan ^{-1} x=t \Rightarrow \frac{1}{1+x^2}=\frac{d t}{d x} \Rightarrow d x=\left(1+x^2\right) d t$
$\therefore \int \frac{e^{\operatorname{lan}^{-1} x}}{1+x^2} d x=\int \frac{e^t}{1+x^2}\left(1+x^2\right) d t$
$=\int e^t d t=e^t+C=e^{\tan ^{-1} x}+C$
D. $\int \frac{e^{2 x}-1}{e^{2 x}+1} d x=\int \frac{\left(e^x \cdot e^x-1\right)}{\left(e^x \cdot e^x+1\right)} d x$
$-\int \frac{e^x\left(e^x-\frac{1}{e^x}\right)}{e^x\left(e^x+\frac{1}{e^x}\right)} d x-\int \frac{\left(e^x-e^{-x}\right)}{\left(e^x+e^{-x}\right)} d x$
Let $e^x+e^{-x}=t$
$\Rightarrow e^x-e^{-x}=\frac{d t}{d x}$
$\Rightarrow d x=\frac{d t}{e^x-e^{-x}}$
$ \Rightarrow \int \frac{e^{2 x}-1}{e^{2 x}+1} d x =\int \frac{\left(e^x-e^{-x}\right)}{t} \frac{d t}{e^x-e^{-x}} $
$ =\int \frac{1}{t} d t=\log |t|+C $
$ =\log \left|e^x+e^{-x}\right|+C$