If $\int \frac{e^x-1}{e^x+1} d x=f(x)+c$, then $f(x)$ is equal to
If $\int \frac{e^x-1}{e^x+1} d x=f(x)+c$, then $f(x)$ is equal to
$2 \log \left(e^x+1\right)$
$\log \left(e^{2 x}-1\right)$
$2 \log \left(e^x+1\right)-x$
$\log \left(e^{2 x}+1\right)$
Solution
We have, $\int \frac{e^x-1}{e^x+1} d x$
$=\int\left(\frac{2 e^x}{e^x+1}-1\right) d x$
$=2 \log \left(e^x+1\right)-x+c$
On comparing with $f(x)+c$, we get $f(x)=2 \log \left(e^x+1\right)-x$