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