How to integrate $\int_0^{\infty} \frac{x}{e^x+1} dx$

Alternatively, let $t=e^{-x}$ to express the integral as

$$\int_0^{\infty} \frac{x}{e^x+1} dx= - \int_0^1 \frac{\ln t}{1+t}dt \overset{IBP} = \int_0^1 \frac{\ln (1+t)}{t}dt = \frac{\pi^2}{12} $$ where the result Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$ is used.

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 : $$ \int_{0}^{+\infty}{\frac{x}{\mathrm{e}^{x}+1}\,\mathrm{d}x}=\eta\left(2\right)=\left(1-\frac{1}{2}\right)\zeta\left(2\right)=\frac{\pi^{2}}{12} $$