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

$$\begin{gathered} \begin{array}{l} \int \frac{e^x-1}{e^x+1}dx\\ =\int \frac{e^x}{e^x+1}dx-\int \frac{1}{e^x+1}dx\\ =\int \frac{1}{e^x+1}d\left(e^x+1\right)+\int \frac{1}{e^{-x}+1}d\left(e^{-x}+1\right)\end{array} \\ \begin{array}{l}=\log \left(e^x+1\right)+\log \left(e^{-x}+1\right)+C\\ =\log \left(e^x+1\right)+\log \left(\frac{e^x+1}{e^x}\right)+C\\ =\log {\left(e^x+1\right)}^2-\log e^x+C=2\log \left(e^x+1\right)-x+C\end{array} \end{gathered}$$