answer from book $x - \ln(e^x + 1)$ $~~~$ *I get * $- \ln \vert 1 + e^{-x} \vert + C $
$$\int \frac{e^{-x}}{1+e^{-x}}\, \mathrm{d}x $$
$$\begin{array}{l c l} u & = & 1 + e^{-x} &\\ \mathrm{d}u & = & -e^{-x}\: \mathrm{d}x \end{array} $$
$$-\int \frac{\mathrm{d}u}{u} $$
$$\Rightarrow -\ln \vert 1 + e^{-x} \vert + C $$
I don't think I made any mistake. How do I get the answer in the book?