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.