Mathematics

Integrate $$\displaystyle \int\dfrac{e^{x}-1}{e^{x}+1}dx$$


ANSWER

$$2\log(e^{-x}+1)-x+C$$


SOLUTION
$$\int_{}^{} {\dfrac{{{e^x} - 1}}{{{e^x} + 1}}dx} $$
putting $${e^x} + 1 = t$$
         $$ \Rightarrow {e^x}dx = dt$$
        $$ \Rightarrow dx = \dfrac{{dt}}{{{e^x}}}$$
         $$ \Rightarrow dx = \dfrac{{dt}}{{t - 1}}$$
$$ = \int_{}^{} {\dfrac{{t - 2}}{{t\left( {t - 1} \right)}}dt} $$
$$ = \int_{}^{} {\left( {\dfrac{2}{t} - \dfrac{1}{{t - 1}}} \right)dt} $$
$$ = 2\log t - \log \left( {t - 1} \right) + c$$
$$ = 2\log \left( {{e^x} + 1} \right) - \log \left( {{e^x}} \right) + c$$
$$ = 2\log \left( {{e^x} + 1} \right) - x + c$$
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Single Correct Medium Published on 17th 09, 2020
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