Improper integral of $\frac{x}{e^{x}+1}$ The improper integral of $\frac{x}{e^x-1}$ (along the positive real line) comes up in a lot of places, you can even invoke the Riemann-zeta and Gamma functions to solve it nicely. 
However, I just decided to look at $\int_0^\infty \frac{x}{e^x+1}dx$, to see if it would also be interesting. It has a nice solution, that I have no satisfying way to get (taking a value of dilogarithm). I'm wondering if anyone has a slick way to do it. Residues? Some slick series argument? 
 A: Putting Dirichlet eta function and Gamma function at work, we get
$$\eta(s)*\Gamma(s)=\int_0^{\infty}\frac{x^{s-1}}{e^x+1}\mathrm{dx},\space Re(s)>1$$
and by letting $s=2$, we obtain
$$\int_0^{\infty}\frac{x}{e^x+1}\mathrm{dx}=\eta(2)*\Gamma(2)=\frac{\pi^2}{12}$$
Chris.
A: $$\int_0^{\infty} \dfrac{x}{e^x+1} dx = \int_0^{\infty} \dfrac{x e^{-x}}{1+e^{-x}} dx = \int_0^{\infty} xe^{-x} \left(\sum_{k=0}^{\infty} (-1)^k e^{-kx} \right) dx$$
Now $$\int_0^{\infty} xe^{-(m+1)x} dx = \dfrac1{(m+1)^2}$$
Hence,
$$\int_0^{\infty} xe^{-x} \left(\sum_{k=0}^{\infty} (-1)^k e^{-kx} \right) dx = \sum_{k=0}^{\infty} (-1)^k \left(\int_0^{\infty} x e^{-(k+1)x} dx \right) = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1)^2} = \dfrac{\pi^2}{12}$$

EDIT
In general, for $n$ in the right half of the complex plane, i.e. $\text{Real}(n) > 0$,
$$\int_0^{\infty} \dfrac{x^n}{e^x+1} dx = \int_0^{\infty} \dfrac{x^n e^{-x}}{1+e^{-x}} dx = \int_0^{\infty} x^ne^{-x} \left(\sum_{k=0}^{\infty} (-1)^k e^{-kx} \right) dx$$
Now $$\int_0^{\infty} x^ne^{-(m+1)x} dx = \dfrac{\Gamma(n+1)}{(m+1)^{n+1}} \,\,\,\,\,\,\, (\text{Follows from definition of }\Gamma(x) \text{ using change of variables})$$
Hence,
\begin{align}
\int_0^{\infty} x^ne^{-x} \left(\sum_{k=0}^{\infty} (-1)^k e^{-kx} \right) dx & = \sum_{k=0}^{\infty} (-1)^k \left(\int_0^{\infty} x^n e^{-(k+1)x} dx \right)\\
& = \Gamma(n+1)\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1)^{n+1}} = \eta(n+1) \Gamma(n+1)
\end{align}
$$\boxed{\displaystyle \color{blue}{\int_0^{\infty} \dfrac{x^n}{e^x+1} dx = \eta(n+1) \Gamma(n+1)}}$$
which is what @Chris's sister has.
A similar technique also reveals the other well-known identity.
$$\boxed{\displaystyle \color{blue}{\int_0^{\infty} \dfrac{x^n}{e^x-1} dx = \zeta(n+1) \Gamma(n+1)}}$$
A: Let
$$
I = \int_0^{ + \infty } {\frac{t}{{e^t  - 1}}dt} 
$$
be the integral that we know well. Then, by a simple change of integration variables,
$$
\frac{I}{2} = \int_0^{ + \infty } {\frac{{2t}}{{e^{2t}  - 1}}dt}.
$$
Thus,
\begin{align*}
\frac{I}{2} & = I - \frac{I}{2} = \int_0^{ + \infty } {\frac{t}{{e^t  - 1}}dt}  - \int_0^{ + \infty } {\frac{{2t}}{{e^{2t}  - 1}}dt} \\ & =\int_0^{ + \infty } {\frac{{t(e^t  + 1) - 2t}}{{e^{2t}  - 1}}dt}  = \int_0^{ + \infty } {\frac{{t(e^t  - 1)}}{{e^{2t}  - 1}}dt}  = \int_0^{ + \infty } {\frac{t}{{e^t  + 1}}dt} .
\end{align*}
