Looks are deceptive!

$\begin{aligned}\int_0^\infty \frac x{e^x-1}dx &=\int_0^\infty {xe^{-x}}\frac1{1-e^{-x}}dx\\ &=\int_0^\infty {xe^{-x}}\sum_{n=0}^\infty \left(e^{-x}\right)^n\ dx\\ &\text{Note that }0<e^{-x}<1\text{ for }x>0\text{, so this is within the interval of convergence for the sum.}\\ &=\int_0^\infty \sum_{n=0}^\infty xe^{-x}e^{-nx}\ dx\\ &=\int_0^\infty \sum_{n=0}^\infty xe^{-(n+1)x}dx\\ &=\int_0^\infty \sum_{n=1}^\infty xe^{-nx}dx\\ &\text{Consider the integral of a single element of this sum, we can solve using integration by parts:}\\ \int_0^\infty xe^{-nx}dx &=\left[\frac xne^{-nx}\right]_0^\infty\ \ -\ \ \int_0^\infty \frac 1ne^{-nx}dx\\ &=\left(\lim_{x\to\infty} \frac x{ne^{nx}}\right) -0\ \ -\ \ \left[\frac1{n^2}e^{-nx}\right]_0^\infty\\ &=0\ \ -\ \ \left(\lim_{x\to\infty} \frac1{n^2e^{nx}}\right) + \frac1{n^2}\\ &=-0+\frac1{n^2}=\frac1{n^2}\\ &\text{Since the integral exists for any }n\text{, we can swap the integral and sum:}\\ \int_0^\infty \sum_{n=1}^\infty xe^{-nx}dx &=\sum_{n=1}^\infty \int_0^\infty xe^{-nx}dx\\ &=\sum_{n=1}^\infty \frac1{n^2}\\ &\text{This famous sum evaluates to }\frac{\pi^2}6\text{, and so}\\ &\int_0^\infty \frac x{e^x-1}dx =\frac{\pi^2}6\qquad\blacksquare\end{aligned}$

$\int \frac{x}{e^x-1} \, dx\Rightarrow x \log \left(1-e^{-x}\right)-\text{Li}_2\left(e^{-x}\right)$

The integral is indeterminate at both ends. The integral has limits at both ends.

$\underset{x\to 0}{\text{lim}}\left(x \log \left(1-e^{-x}\right)-\text{Li}_2\left(e^{-x}\right)\right) \Rightarrow -\frac{\pi ^2}{6}$

$\underset{x\to \infty }{\text{lim}}\left(x \log \left(1-e^{-x}\right)-\text{Li}_2\left(e^{-x}\right)\right) \Rightarrow 0$

This gives the result $\frac{\pi ^2}{6}$ .

The answer is $8$ .

$\int_0^{\infty}\frac{x}{e^x-1}dx=\int_0^{\infty}\frac{xe^{-x}}{1-e^{-x}}dx$ . Use substitution $x=-m$ . The new integral becomes $\int_0^{-\infty}\frac{me^m}{1-e^m}dm$ . Use the substitution $1-e^m=u$ . The new integral then becomes $-\int_0^1\frac{\ln\left(1-u\right)}{u}du$ . Note that $\ln\left(1-u\right)\ =\ -\sum_{n=1}^{\infty}\frac{u^n}{n}$ . So this becomes $\int_0^1\sum_{n=1}^{\infty}\frac{u^{n-1}}{n}du$ . Integrating term by term this becomes $\frac{u^n}{n^2}$ . The integral then becomes $\sum_{n=1}^{\infty}\frac{1^n}{n^2}-\sum_{n=1}^{\infty}\frac{0^n}{n^2}$ , which is simply $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ which gives us the answer of $8$