Question

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

• A. $2\log (e^x+1)$
• B. $\log (e^x-1)$
• C. $2\log (e^x+1)-x$
• D. $\log (e^{2x}+1)$

Answer 1

Answer: (C)

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

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

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

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

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

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

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

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