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.

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.
