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


ANSWER

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


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


ANSWER

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


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
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