#### Passage

Let $\displaystyle 2I_{1}+I_{2}=\int \frac {e^{x}}{e^{2x}+e^{-2x}}dx$  and  $\displaystyle I_{1}+2I_{2}=\int \frac {e^{-x}}{e^{2x}+e^{-2x}}dx$
On the basis of above information, answer the following questions :
Mathematics

# $I_{2}$ is equal to

$\displaystyle \frac {1}{6\sqrt{2}}tan^{-1}\left ( \frac {e^{x}-e^{-x}}{\sqrt{2}} \right )-\frac {1}{4\sqrt{2}}\iota n \left | \frac {e^{x}+e^{-x}-\sqrt{2}}{e^{x}+e^{-x}+\sqrt{2}} \right |+C$

##### SOLUTION
$\displaystyle { 2I }_{ 1 }+{ I }_{ 2 }=\int { \frac { { e }^{ x } }{ { e }^{ 2x }+{ e }^{ -2x } } } dx$

Put ${ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt$
$\displaystyle \therefore { 2I }_{ 1 }+{ I }_{ 2 }=\int { \frac { { t }^{ 2 } }{ { t }^{ 4 }+1 } dt }$

$\displaystyle =\int { \left( -\frac { t }{ 2\sqrt { 2 } \left( -{ t }^{ 2 }+\sqrt { 2 } t-1 \right) } -\frac { t }{ 2\sqrt { 2 } \left( { t }^{ 2 }+2\sqrt { 2 } t+1 \right) } \right) } dt$

$\displaystyle =\frac { -1 }{ 2\sqrt { 2 } } \int { \frac { t }{ { t }^{ 2 }+\sqrt { 2+1 } } } dt-\frac { 1 }{ 2\sqrt { 2 } } \int { \frac { t }{ -{ t }^{ 2 }+\sqrt { 2 } t-1 } dt }$

$\displaystyle =-\frac { 1 }{ 2\sqrt { 2 } } \int { \left( \frac { 2t+\sqrt { 2 } }{ 2\left( { t }^{ 2 }+\sqrt { 2 } t+1 \right) } -\frac { 1 }{ \sqrt { 2 } \left( { t }^{ 2 }+\sqrt { 2 } t+1 \right) } \right) dt }$

$\displaystyle -\frac { 1 }{ 2\sqrt { 2 } } \int { \left( \frac { 1 }{ \sqrt { 2 } \left( { t }^{ 2 }+\sqrt { 2 } t+1 \right) } -\frac { \sqrt { 2 } -2t }{ 2\left( -{ t }^{ 2 }+\sqrt { 2 } t-1 \right) } \right) } dt$

$\displaystyle =\frac { \log { \left( -{ t }^{ 2 }+\sqrt { 2 } t-1 \right) } }{ 4\sqrt { 2 } } -\frac { \log { \left( { t }^{ 2 }+\sqrt { 2 } t \right) +1 } }{ 4\sqrt { 2 } } -\frac { \tan ^{ -1 }{ \left( 1-\sqrt { 2 } t \right) } }{ 2\sqrt { 2 } } +\frac { \tan ^{ -1 }{ \left( \sqrt { 2 } t+1 \right) } }{ 2\sqrt { 2 } }$

$\displaystyle =\frac { \log { \left( { e }^{ 2x }+\sqrt { 2 } { e }^{ x }+1 \right) } }{ 4\sqrt { 2 } } -\frac { \log { \left( { e }^{ 2x }+\sqrt { 2 } { e }^{ x }+1 \right) } }{ 4\sqrt { 2 } } -\frac { \tan ^{ -1 }{ \left( 1-\sqrt { 2 } { e }^{ x } \right) } }{ 2\sqrt { z } } +\frac { \tan ^{ -1 }{ \left( \sqrt { 2 } { e }^{ x }+1 \right) } }{ 2\sqrt { 2 } }$

Now $\displaystyle { I }_{ 1 }+{ 2 }I_{ 2 }=\int { \frac { { e }^{ -x } }{ { e }^{ 2x }+{ e }^{ -2x } } } dx$

Put $u={ e }^{ -x }\Rightarrow du=-{ e }^{ -x }dx$
$\displaystyle \therefore { I }_{ 1 }+{ 2I }_{ 2 }=-\int { \frac { { u }^{ 2 } }{ { u }^{ 2 }+1 } } du$

$\displaystyle =-\frac { \log { \left( -{ u }^{ 2 }+\sqrt { 2 } u-1 \right) } }{ 4\sqrt { 2 } } +\log { \frac { \left( { u }^{ 2 }+\sqrt { 2 } u+1 \right) }{ 4\sqrt { 2 } } } +\frac { \tan ^{ -1 }{ \left( 1-\sqrt { 2 } u \right) } }{ 2\sqrt { 2 } } -\frac { \tan ^{ -1 }{ \left( \sqrt { 2 } u-1 \right) } }{ 2\sqrt { 2 } }$

$\displaystyle =-\frac { \log { \left( { -e }^{ -2x }+\sqrt { 2 } { e }^{ -x }-1 \right) } }{ 4\sqrt { 2 } } +\frac { \log { \left( { e }^{ -2x }+\sqrt { 2 } { e }^{ -x }+1 \right) } }{ 4\sqrt { 2 } } +\frac { \tan ^{ -1 }{ \left( 1-\sqrt { 2 } { e }^{ -x } \right) } }{ 2\sqrt { 2 } } +\frac { \tan ^{ -1 }{ \left( \sqrt { 2 } { e }^{ -x }+1 \right) } }{ 2\sqrt { 2 } }$

Therefore $\displaystyle { I }_{ 2 }=\frac { 1 }{ 6\sqrt { 2 } } \tan ^{ -1 }{ \frac { { e }^{ x }-{ e }^{ -x } }{ \sqrt { z } } } -\frac { 1 }{ 4\sqrt { 2 } } \log { \left| \frac { { e }^{ x }+{ e }^{ -x }-\sqrt { 2 } }{ { e }^{ x }+{ e }^{ -x }+\sqrt { 2 } } \right| +c }$

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Single Correct Hard Published on 17th 09, 2020
Mathematics

# $I_{1}$ is equal to

$\displaystyle \frac {1}{6\sqrt{2}}tan^{-1}\left ( \frac {e^{x}-e^{-x}}{\sqrt{2}} \right )+\frac {1}{4\sqrt{2}}\iota n \left | \frac {e^{x}+e^{-x}-\sqrt{2}}{e^{x}+e^{-x}+\sqrt{2}} \right |+C$

##### SOLUTION
$\displaystyle 3I_{1}+3I_{2}= \int \frac{e^{x}+e^{-x}}{(e^{x}-e^{-x})^{2}+2}dx$
Let, $e^{x}-e^{-x}=t$  $\Rightarrow (e^{x}+e^{-x})dx=dt$
$\therefore \displaystyle 3I_{1}+3I_{2}= \int \frac{dt}{t^{2}+2}=\frac{1}{\sqrt{2}}tan^{-1}\left ( \frac{t}{\sqrt{2}} \right )+C$
$\displaystyle I_{1}+I_{2}= \frac{1}{3\sqrt{2}}tan^{-1}\left ( \frac {e^{x}-e^{-x}}{\sqrt{2}} \right )+C$   .......$(i)$
&    $\displaystyle I_{1}-I_{2}= \int \frac {e^{x}-e^{-x}}{(e^{x}+e^{-x})^{2}-2}dx$
Let, $e^{x}+e^{-x}=z \Rightarrow (e^{x}-e^{-x})dx=dz$
$\therefore \displaystyle I_{1}-I_{2}= \int \frac {dz}{z^{2}-2}=\frac {1}{2\sqrt{2}}\iota n \left | \frac{z-\sqrt{2}}{z+\sqrt{2}} \right |+C$   .......$(ii)$
From $(i)$  &  $(ii)$
$\displaystyle I_{1}= \frac {1}{6\sqrt{2}}tan^{-1}\left ( \frac {e^{x}-e^{-x}}{\sqrt{2}}\right)+\frac {1}{4\sqrt{2}}\iota n \left | \frac{e^{x}+e^{-x}-\sqrt{2}}{e^{x}+e^{-x}+\sqrt{2}} \right |+C$
&
$\displaystyle I_{2}= \frac {1}{6\sqrt{2}}tan^{-1}\left ( \frac {e^{x}-e^{-x}}{\sqrt{2}}\right)-\frac {1}{4\sqrt{2}}\iota n \left | \frac{e^{x}+e^{-x}-\sqrt{2}}{e^{x}+e^{-x}+\sqrt{2}} \right |+C$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Single Correct Medium
Area enclosed by $x^2 = 4y$ and $y=\frac{8}{x^2 + 4}$ is
• A. $\pi - \frac{2}{3}$
• B. $\pi - \frac{1}{3}$
• C. None of these
• D. $2\pi - \frac{4}{3}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int \frac{1}{\left ( x+1 \right )\sqrt{x^2-1}}dx$
• A. $\displaystyle \sqrt{\frac{x-1}{x^2+1}}$
• B. $\displaystyle \sqrt{\frac{x+1}{x-1}}+C$
• C. $\displaystyle \sqrt{\frac{x+1}{x-1}}$
• D. $\displaystyle \sqrt{\frac{x-1}{x+1}}+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\displaystyle I_{1}= \int_{-4}^{-5}e^{\left ( x +5 \right )^{2}}dx$ and $\displaystyle I_{2}= 3\int_{{1}/{3}}^{{2}/{3}}e^{\left ( 3x -2 \right )^{2}}dx$

then $I_{1}+I_{2}$ equals?
• A. $\displaystyle \frac{1}{3}$
• B. $\displaystyle -\frac{1}{3}$
• C. None of these
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Evaluate $\displaystyle \int \frac{\log\left ( x+\sqrt{1+x^{2}} \right )}{\sqrt{1+x^{2}}}dx.$

• A. $\displaystyle \frac{1}{2\left [ \log\left ( x+\sqrt{1+x^{2}} \right ) \right ]^{2}} +c$
• B. $\displaystyle \frac{\left[\log\left ( x+\sqrt{1+x^{2}} \right )\right]^2}{4}+c$
• C. $\displaystyle \frac{\left[\log\left ( x-\sqrt{1+x^{2}} \right )\right]^2}{2}+c$
• D. $\displaystyle \frac{\left[\log\left ( x+\sqrt{1+x^{2}} \right )\right]^2}{2}+c$

$\frac{1}{2}\int {\frac{{2x}}{{{{\left( {1 + x} \right)}^2}}}dx}$