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

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

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

$=\log |e^x+1|-\int \frac{(e^x+1)-e^x}{e^x+1}dx$

$=\log |e^x+1|-\int \left(1-\frac{e^x}{e^x+1}\right)dx$

$=\log (e^x+1)-x+\log (e^x+1)+c$

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