The improper integral $$\int_0^\infty \frac{x}{e^{x}-1}dx$$ comes up in a lot of places. One 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?
$$\int_0^{\infty} \dfrac{x}{e^x+1} dx = \int_0^{\infty} \dfrac{x e^{-x}}{1+e^{-x}} dx = \int_0^{\infty} xe^{-x} \left(\sum_{k=0}^{\infty} (-1)^k e^{-kx} \right) dx$$ 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. $$\boxed{\displaystyle \color{blue}{\int_0^{\infty} \dfrac{x^n}{e^x-1} dx = \zeta(n+1) \Gamma(n+1)}}$$
The twin integrals $$I_{\pm} = \int_0^{\infty}\frac{x}{e^x\pm1}{dx} $$ are handled at once via the elementary procedure below. \begin{align} I_- - I_+ &= \int_0^{\infty}\frac{2x}{e^{2x}-1}\overset{2x\to x}{dx}=\frac12I_-\\ I_- + I_+ &=\int_0^\infty\frac{2xe^x}{e^{2x}-1}dx =\int_0^\infty\int_0^1 \frac{2t e^x}{1+(e^{2x}-1)t^2}dt\ dx\\ &=\int_0^1 \frac{\arccos t}{\sqrt{1-t^2}}dt=-\arccos^2t\bigg|_0^1=\frac{\pi^2}{4} \end{align} which yields $I_+=\frac{\pi^2}{12}$ and $I_-=\frac{\pi^2}{6}$.
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.
In fact, $$ I = \int_0^{ + \infty } {\frac{t}{{e^t + 1}}dt}=-\int_0^\infty t\;d\ln(1+e^{-t})=-t\ln(1+e^{-t})\bigg|_0^\infty+\int_0^\infty\ln(1+e^{-t})dt=\int_0^1\frac{\ln(1+t)}{t}dt=\frac{\pi^2}{12}. $$
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*}
I've noticed no one has attempted solving the indefinite integral, rather using identities and other functions on the bounds $0$ to $\infty$. Here is a method to solve the indefinite integral and evaluate the integral over any bounds in the domain. I wouldn't consider this a very elegant method, but still a solution nonetheless. I began by substituting $u=e^x$ giving us: $$\int\frac{\ln(u)}{u(u+1)}du$$ Now using partial fraction decomposition: $$\int\frac{\ln(u)}{u}du-\int\frac{\ln(u)}{u+1}du$$ You'll notice we can substitute the dilogarithm integral: $\text{Li}_2(u)=\int-\frac{\ln(1-u)}{u}du$
After solving both integrals we get: $$-\ln(u)\ln(u+1)+\frac{\ln^2(u)}{2}0-\text{Li}_2(-u)$$ Now we undo our substitution and have completed the indefinite integral. $$-x \ln\left(\mathrm{e}^{x} + 1\right) - \operatorname{Li}_{2}\left(-\mathrm{e}^{x}\right) + \frac{x^{2}}{2}+C$$