Find $A =\int \frac{x \, dx}{e^x +1 }$ 
Find $\displaystyle  A =\int \frac{x \, dx}{e^x +1 }$

Using Wolfram Alpha this yields  $-\operatorname{Li}_2(-e^x) + \frac{x^2}{2}-x \ln(e^x +1)$, where $\operatorname{Li}$ is the polylogaritm function. I am pretty really new to polylogarithms but I wanted to know how WA arrived at this result and if possible if there are any books that could help me grasp the use of these special functions in integration.
 A: Note that 
$$\begin{align}
\int\,\frac{x\,\exp(x)}{\exp(x)+1}\,\text{d}x&=\int\,x\,\text{d}\ln\big(\exp(x)+1\big)\\&=x\,\ln\big(\exp(x)+1\big)-\int\,\ln\big(\exp(x)+1\big)\,\text{d}x\,.
\end{align}$$
Then, we have that
$$\int\,\ln\big(\exp(x)+1\big)\,\text{d}x=\int\,\frac{\ln\Big(1-\big(-\exp(x)\big)\Big)}{\big(-\exp(x)\big)}\,\text{d}\big(-\exp(x)\big)=-\text{Li}_2\big(-\exp(x)\big)+C\,,$$
where $C$ is a constant (as $\text{Li}_2(z)=-\int_0^z\,\frac{\ln(1-t)}{t}\,\text{d}t$, according to this).
Thus,
$$\begin{align}\int\,\frac{x}{\exp(x)+1}\,\text{d}x
&=\int\,\left(x-\frac{x\,\exp(x)}{\exp(x)+1}\right)\,\text{d}x
\\&=\frac{1}{2}x^2-x\,\ln\big(\exp(x)+1\big)-\text{Li}_{2}\big(-\exp(x)\big)+C\,.\end{align}$$
You can prove $\text{Li}_2(z)=-\int_0^z\,\frac{\ln(1-t)}{t}\,\text{d}t$ directly via
$$\begin{align}\text{Li}_2(z)&=\sum_{k=1}^\infty\,\frac{z^k}{k^2}=\sum_{k=1}^\infty\,\int_0^z\frac{t^{k-1}}{k}\,\text{d}t=\int_0^z\,\frac{1}{t}\,\sum_{k=1}^\infty\,\frac{t^k}{k}\,\text{d}t
\\
&=\int_0^z\,\frac{-\ln(1-t)}{t}\,\text{d}t\,,
\end{align}$$
for all $z\in\mathbb{C}$ with $|z|<1$.  Then, analytic continuation handles the rest.
A: Since $\int xe^{-nx}\,\mathrm{d}x=-\frac1{n^2}(nx+1)e^{-nx}+C$,
$$
\begin{align}
\int \frac{x\,\mathrm{d}x}{e^x+1}
&=\int x\left(e^{-x}-e^{-2x}+e^{-3x}+\dots\right)\mathrm{d}x\tag{1}\\
&=C-(x+1)e^{-x}+\tfrac14(2x+1)e^{-2x}-\tfrac19(3x+1)e^{-3x}+\dots\tag{2}\\
&=C-x\left(e^{-x}-\tfrac12e^{-2x}+\tfrac13e^{-3x}-\dots\right)+\left(-e^{-x}+\tfrac14e^{-2x}-\tfrac19e^{-3x}+\dots\right)\tag{3}\\
&=C-x\log\left(1+e^{-x}\right)+\mathrm{Li}_2\left(-e^{-x}\right)\tag{4}\\
&=C+x^2-x\log\left(e^x+1\right)-\tfrac{\pi^2}6-\mathrm{Li}_2\!\left(-e^x\right)-\tfrac12x^2\tag{5}\\
&=C_2+\tfrac12x^2-x\log\left(e^x+1\right)-\mathrm{Li}_2\!\left(-e^x\right)\tag{6}
\end{align}
$$
Explanation:
$(1)$: write $\frac1{e^x+1}$ as a geometric series
$(2)$: use the formula from above
$(3)$: separate terms
$(4)$: use the series for $\log(1+x)$ and $\mathrm{Li}_2(x)$
$(5)$: use $(5)$ from this answer
$(6)$: collect terms
A: \begin{align}
I&=\int\frac{x\ dx}{e^x+1}=\int\frac{xe^{-x}}{1+e^{-x}}\ dx\overset{e^{-x}=y}{=}\int\frac{\ln y}{1+y}\ dy\\
&\overset{IBP}{=}\ln(1+y)\ln y-\int\frac{\ln(1+y)}{y}\ dy\\
&=\ln(1+y)\ln y-(-\operatorname{Li}_2(-y))\\
&=\ln(1+e^{-x})\ln(e^{-x})+\operatorname{Li}_2(-e^{-x})\\
&\{\color{red}{\text{use }\operatorname{Li}_2(-1/x)=-\operatorname{Li}_2(-x)-\frac12\ln^2x-\zeta(2)}\}\\
&=-x\ln\left(\frac{e^x+1}{e^x}\right)-\operatorname{Li}_2(-e^{x})-\frac12\ln^2(e^x)-\zeta(2)\\
&=-x\ln(1+e^x)+x^2-\operatorname{Li}_2(-e^{x})-\frac12x^2-\zeta(2)\\
&=\frac12x^2-x\ln(1+e^x)-\operatorname{Li}_2(-e^{x})-\zeta(2)+C
\end{align}

Note: The identity in red is proved here but just replace $x$ with $-x$.
