How to integrate $\int_0^{\infty} \frac{x}{e^x+1} dx$ I want to integrate it explicitly. I looked up if this had an exact solution, and I got that it is $\frac{\pi^2}{12}$ here: https://www.integral-calculator.com/#expr=x%2F%28e%5Ex%2B1%29&lbound=0&ubound=inf
I thought it could be solved using geometric series, so I tried to turn my integral into something similar:
$$\int_0^{\infty} \frac{x}{e^x+1} dx= \int_0^{\infty} \frac{x}{e^x+1} \frac{e^x-1}{e^x-1} dx=\int_0^{\infty} \frac{x e^x }{e^{2x}-1}dx -\int_0^{\infty} \frac{x}{e^{2x}-1} dx= \int_0^{\infty} \frac{x e^x }{e^{2x}-1}dx -\frac{1}{4}\int_0^{\infty}  \frac{x}{e^{x}-1} dx=\int_0^{\infty} \frac{x e^x}{e^{2x}-1} - \frac{1}{4}\frac{\pi^2}{6}$$
Where the second term is easy to solve with geometric series, and it's also defined with the Riemann zeta:
$$\int_0^{\infty} \frac{x}{e^x-1}=\zeta(2)=\frac{\pi^2}{6}$$
I think I can use geometric series also with the other integral, I've tried to solve it this way and I got:
$$\int_0^{\infty}\frac{xe^x}{e^x-1}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{\left(n-\frac{1}{2}\right)^2} $$
And I don't know how to obtain a value from this infinite sum, although if I'm not wrong it should be $\pi^2/2$.
Anyone knows how to do this? Also, I think I might be looping the loop and there's some easier way to proceed from the beggining, so I'd appreciate another point of view to solve this.
Thanks.

Let $ n\in\mathbb{N} $, we have : \begin{aligned}\left\vert\sum_{k=1}^{n}{\left(-1\right)^{k-1}\int_{0}^{+\infty}{x\,\mathrm{e}^{-kx}\,\mathrm{d}x}}-\int_{0}^{+\infty}{\frac{x}{\mathrm{e}^{x}+1}\,\mathrm{d}x}\right\vert&=\left\vert\int_{0}^{+\infty}{x\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k-1}\,\mathrm{e}^{-kx}}\,\mathrm{d}x}\right\vert\\ &\leq\int_{0}^{+\infty}{x\left\vert\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k-1}\,\mathrm{e}^{-kx}}\right\vert\mathrm{d}x}\\ &\leq\int_{0}^{+\infty}{x\,\mathrm{e}^{-\left(n+1\right)x}\,\mathrm{d}x}=\frac{\Gamma\left(1\right)}{\left(n+1\right)^{2}}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
Which means that : $$ \sum_{n=1}^{+\infty}{\frac{\left(-1\right)^{n-1}}{n^{2}}}=\lim_{n\to +\infty}{\sum_{k=1}^{n}{\left(-1\right)^{k-1}\int_{0}^{+\infty}{x\,\mathrm{e}^{-kx}\,\mathrm{d}x}}}=\int_{0}^{+\infty}{\frac{x}{\mathrm{e}^{x}+1}\,\mathrm{d}x} $$
Thus : $$ \int_{0}^{+\infty}{\frac{x}{\mathrm{e}^{x}+1}\,\mathrm{d}x}=\eta\left(2\right)=\left(1-\frac{1}{2}\right)\zeta\left(2\right)=\frac{\pi^{2}}{12} $$
