Mathematics

If, $\displaystyle \int \dfrac{e^x - 1}{e^x + 1} dx = f(x) + c$, then $f(x)$ is equal to

$2 \, \log (e^x + 1) - x$

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

let $u = e^x ; du = e^x dx$

$=\int \dfrac{u-1}{u(u+1)}du$

$\int \dfrac {2u-(u+1)} {u(u+1)} du$

$=2\int \dfrac{1}{u+1}du -\int \dfrac{1}{u} du$

$=2\ln (u+1)-\ln (u)+C$

$=2 \ln (e^x+1)-x+C$

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

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