How to solve this integral: $\int_{-\infty}^\infty\frac{x^2 e^x}{(e^x+1)^2}\:dx$ I am trying to solve an integral like this:
$$
I=\int \frac{x^2 e^x}{(e^x+1)^2} dx
$$
And I get this answer:
$$
\int \frac{x^2 e^x}{(e^x+1)^2}dx=x^2-\frac{x^2}{e^x+1}-2x\text{ln}(e^x+1)+2\int\text{ln}(e^x+1)dx
$$
Where ln denotes natural logarithm.
The last integral can be rewritten as Li$_2(-e^x)$:
\begin{gather}
\int\text{ln}(e^x+1)dx=\begin{cases} u=e^x\\
du=e^xdx\Rightarrow dx={du\over u}\end{cases}\Rightarrow\\
\Rightarrow \int\text{ln}(e^x+1)dx=\int\frac{\text{ln}(u+1)}{u} du
\end{gather}
As Li$_1(z)=-\text{ln}(1-z)\Rightarrow \text{ln}(u+1)=\text{ln}(1-(-u))=-\text{Li}_1(-u)$, it yields:
\begin{gather}
\int\frac{\text{ln}(u+1)}{u} du=-\int\frac{\text{Li}_1(-u)}{u}du=-\text{Li}_2(-u)=-\text{Li}_2(-e^x)
\end{gather}
Finally:
\begin{gather}
\int \frac{x^2 e^x}{(e^x+1)^2} dx=\int \frac{x^2 e^x}{(e^x+1)^2}dx=x^2-\frac{x^2}{e^x+1}-2x\text{ln}(e^x+1)-2\text{Li}_2(-e^x)
\end{gather}
Now, I need to evaluate $I$ from $-\infty$ to $\infty$. I have solved it in Wolfram Alpha, but although it often shows the step by step option, in this case, it doesn't. The result is $\pi^2/3$, which is barely $\zeta(2)$, where $\zeta(\cdot)$ is the Riemann's zeta function, closely related to polylogarithms. So the question is how to evaluate this integral from $-\infty$ to $\infty$. Thank you in advance.
 A: Hint. Observe that we have
$$
\begin{align}
\int_{-\infty}^\infty \frac{x^2 e^x}{(e^x+1)^2}\:dx&=\int_{-\infty}^0 \frac{x^2 e^x}{(e^x+1)^2}\:dx+\int_0^\infty \frac{x^2 e^x}{(e^x+1)^2}\:dx\\\\
&=2\int_0^\infty \frac{x^2 e^x}{(e^x+1)^2}\:dx\\\\
&=2\int_0^\infty \frac{x^2e^{-x} }{(1+e^{-x})^2}\:dx\\\\
&=2\sum_{n=1}^{\infty}n(-1)^{n-1}\underbrace{\int_0^\infty x^2e^{-nx} \:dx}_\color{red}{{\large 2/n^3}}\\\\
&=4\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\\\\
&=\frac{\pi^2}3.
\end{align}
$$


Some details. One may recall that (the standard geometric sum)
  $$
\sum_{n=0}^{\infty}(-1)^{n}u^n=\frac{1}{1+u}, \qquad |u|<1,
$$ then differentiating term-wise and multiplying by $u$ gives
  $$
\sum_{n=1}^{\infty}n(-1)^{n-1}u^n=\frac{u}{(1+u)^2}, \qquad |u|<1.
$$ By putting $u:=e^{-x}$ in the preceding identity, we get
  $$
\frac{e^{-x}}{(1+e^{-x})^2}=\sum_{n=1}^{\infty}n(-1)^{n-1}e^{-nx}, \qquad x>0,
$$ leading to 
  $$
\int_0^\infty \frac{x^2e^{-x} }{(1+e^{-x})^2}\:dx=\sum_{n=1}^{\infty}n(-1)^{n-1}\int_0^\infty x^2e^{-nx}\:dx.
$$

A: Often, with such complicated primitives, I suggest not to calculate the primitive, but to use some trick, like introducing a parameter and differentiate with respect to it. Read below if you are interested in such a way to calculate this integral. Tell me if this is far from what you looked for.
First, since the integrand is even, your integral equals
$$
2\int_0^{+\infty}\frac{x^2e^x}{(1+e^x)^2}\,dx.
$$
Introduce
$$
f(a)=\int_0^{+\infty}\frac{x}{1+e^{ax}}\,dx.
$$
Then, your integral equals $-2f'(1)$. To calculate $f(a)$, we note that
$$
\frac{x}{1+e^{ax}}=x\bigl(e^{-ax}-e^{-2ax}+e^{-3ax}-\cdots\bigr)
$$
and
$$
\int_0^{+\infty}(-1)^{k+1}xe^{-akx}\,dx=\frac{1}{a^2}(-1)^{k+1}\frac{1}{k^2}.
$$
We get that
$$
f(a)=\frac{1}{a^2}\sum_{k=1}^{+\infty}(-1)^{k+1}\frac{1}{k^2}=\frac{1}{a^2}\frac{\pi^2}{12}.
$$
It remains to check that
$$
-2f'(1)=-2\cdot(-2)\cdot\frac{\pi^2}{12}=\frac{\pi^2}{3}.
$$
Edit
There are many ways of calculating $\sum_{k=1}^{+\infty}(-1)^{k+1}1/k^2$. One is by using the Basel problem 
$$
\frac{\pi^2}{6}=\sum_{k=1}^{+\infty}\frac{1}{k^2}=-\int_0^1\frac{\ln x}{1-x}\,dx
$$ 
together with (I leave the calculations to you)
$$
\sum_{k=1}^{+\infty}(-1)^{k+1}\frac{1}{k^2}=\int_0^1\frac{\log(1+x)}{x}\,dx=-\frac{1}{2}\int_0^1\frac{\ln x}{1-x}\,dx.
$$
Another is by inserting $x=0$ in the Fourier series (for $x=0$ the $\sim$ below can be replaced by $=$ by a well-known theorem on convergence of Fourier series)
$$
x^2\sim \frac{\pi^2}{3}+4\sum_{k=1}^{+\infty}(-1)^k\frac{\cos kx}{k^2}.
$$
There must be questions on the "alternating Basel problem" on this site, but my search did not find it. Good luck, and tell me if something is still unclear.
A: Because $\frac{x^2e^x}{(e^x+1)^2}=\frac{x^2}{\left(e^{x/2}+e^{-x/2}\right)^2}$ is even, we can integrate by parts and use the Dirichlet eta function:
$$
\begin{align}
\int_{-\infty}^\infty\frac{x^2e^x}{(e^x+1)^2}\,\mathrm{d}x
&=2\int_0^\infty\frac{x^2e^x}{(e^x+1)^2}\,\mathrm{d}x\\
&=-2\int_0^\infty x^2\,\mathrm{d}\frac1{e^x+1}\\
&=4\int_0^\infty\frac{x}{e^x+1}\,\mathrm{d}x\\
&=4\Gamma(2)\,\eta(2)\\[6pt]
&=2\zeta(2)\\
&=\frac{\pi^2}3
\end{align}
$$
