Question

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

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

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