Mathematics

$$\displaystyle \int \frac{e^{x}-1}{e^{x}+1}dx.$$


ANSWER

$$\displaystyle 2\log \left ( e^{x}+1 \right )-\log e^{x}$$


SOLUTION
Let $$ \displaystyle I=\int  \frac { e^{ x }-1 }{ e^{ x }+1 } dx=\int  \frac { e^{ x }\left( e^{ x }-1 \right)  }{ e^{ x }\left( e^{ x }+1 \right)  } dx$$

Put $$ \displaystyle e^{ x }=t\Rightarrow e^{ x }dx=dt$$

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

$$ \displaystyle =2\log  \left( t+1 \right) -\log  t=2\log  \left( e^{ x }+1 \right) -\log  e^{ x }$$
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Single Correct Medium Published on 17th 09, 2020
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