Question

Let $I=\int \frac{e^x}{e^{4x}+1}dx$ and $J=\int \frac{e^{-x}}{e^{-4x}+1}dx$
Then for any arbitrary constant $C$, match the following

Column I		Column II
A	$I$	P	$\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{e^{2x}-1}{\sqrt{2}{e}^x}\right)+C$
B	$J+I$	Q	$\frac{1}{2\sqrt{2}}\ln \left(\frac{e^{2x}-\sqrt{2}{e}^x+1}{e^{2x}+\sqrt{2}{e}^x+1}\right)+C$
C	$J-I$	R	$\frac{1}{2\sqrt{2}}\left({\tan }^{-1}\left(\frac{e^{2x}-1}{\sqrt{2}{e}^x}\right)-\frac{1}{2}\ln \left(\frac{e^{2x}-\sqrt{2}{e}^x+1}{e^{2x}+\sqrt{2}{e}^x+1}\right)\right)+C$
		S	$\frac{1}{2\sqrt{2}}\left({\tan }^{-1}\left(\frac{e^{2x}-1}{\sqrt{2}{e}^x}\right)+\frac{1}{2}\ln \left(\frac{e^{2x}-\sqrt{2}{e}^x+1}{e^{2x}+\sqrt{2}{e}^x+1}\right)\right)+C$

• A. (A) R ; (B) P ; (C) Q
• B. (A) R ; (B) Q ; (C) Q
• C. (A) R ; (B) R ; (C) Q
• D. (A) R ; (B) S ; (C) Q

Answer 1

Answer: (A)

Given $I=\int \frac{e^x}{1+e^{4x}}dx$ and $J=\int \frac{e^{-x}}{1+e^{-4x}}dx$
$J=\int \frac{e^{4x}\cdot e^{-x}}{1+e^{4x}}dx=\int \frac{e^{3x}dx}{1+e^{4x}}$
$\therefore I+J=\int \frac{e^x\left(1+e^{2x}\right)}{1+e^{4x}}dx$
$\text{ put }{e}^x=t\Rightarrow e^{x}dx=dt$
$\therefore I+J=\int \frac{1+t^2}{1+t^4}dt=\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt=\int \frac{1+\frac{1}{t^2}}{{\left(t-\frac{1}{t}\right)}^2+2}dt$
put $t-\frac{1}{t}=y\Rightarrow y=\left(t+\frac{1}{t^2}\right)dt=dy$
$=\int \frac{dy}{y^2+2}=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{y}{\sqrt{2}}=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{t^2-1}{\sqrt{2}t}$
$J+I=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{e^{2x}-1}{\sqrt{2}{e}^x}\right)+C\stackrel{(}{\to }P)$
$\Vert$ ly $J-I=\int \frac{e^x\left(-1+e^{2x}\right)}{1+e^{4x}}dx=\int \frac{-1+t^2}{1+t^4}dt=+\int \frac{-1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt=+\int \frac{-1+\frac{1}{t^2}}{{\left(t+\frac{1}{t}\right)}^2-2}dt\left(t+\frac{1}{t}=y\right)$
$\therefore J-I=+\int \frac{dy}{y^2-2}=-\frac{1}{2\sqrt{2}}\ln \frac{y-\sqrt{2}}{y+\sqrt{2}}$
$J-I=-\frac{1}{2\sqrt{2}}\ln \frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}=\frac{1}{2\sqrt{2}}\cdot \frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1}=\frac{1}{2\sqrt{2}}\cdot \frac{e^{2x}-\sqrt{2}{e}^x+1}{e^{2x}+\sqrt{2}{e}^x+1}\dots (2)\Rightarrow (Q)$
(1) - (2) gives
$2I=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{e^{2x}-1}{\sqrt{2}{e}^x}-\frac{1}{2\sqrt{2}}\frac{e^{2x}-\sqrt{2}{e}^x+1}{e^{2x}+\sqrt{2}{e}^x+1}$
$I=\frac{1}{2\sqrt{2}}\left[{\tan }^{-1}\frac{e^{2x}-1}{\sqrt{2}{e}^x}-\frac{1}{2}\cdot \frac{e^{2x}-\sqrt{2}{e}^x+1}{e^{2x}+\sqrt{2}{e}^x+1}\right]\Rightarrow (R)$