How find this integral $I=\int_{0}^{+\infty}\frac{x}{1+e^x}dx$ Question:

$$I=\int_{0}^{+\infty}\dfrac{x}{1+e^x}dx$$

I know use
$$I=\int_{0}^{\infty}\dfrac{xe^{-x}}{1+e^{-x}}dx=\sum_{k=0}^{\infty}\int_{0}^{\infty}xe^{-(k+1)x}dx=\sum_{k=1}^{\infty}(-1)^{k-1}\cdot\dfrac{1}{k^2}=\dfrac{\pi^2}{12}$$
someone have other methods? Thank you,Because I want collect more methods,Thank you
 A: Another variation of the method based on using series expansion. First, we make the substitution $e^{-x}=t$, and then integrate by parts:
$$I=\int_0^{+\infty}\frac{x}{1+e^x}dx=-\int_0^1\frac{\ln t}{1+t}dt=-\int_0^1\ln t\,d\ln(1+t)=\int_0^1\frac{\ln (1+t)}{t}dt.$$
Now we use Taylor series for $\ln(1+t)$:
$$I=\int_0^1\frac{1}{t}\sum_{n=1}^\infty(-1)^{n-1}\frac{t^n}{n} dt=\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\int_0^1 t^{n-1} dt=\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n^2}=\frac{\pi^2}{12}.$$
A: Consider $$I=\int\dfrac{x}{1+e^x}dx$$ and change variable $x=\log(y)$. So $$I=\int\dfrac{\log(y)}{y(y+1)}dy=\int\dfrac{\log(y)}{y}dy-\int\dfrac{\log(y)}{y+1}dy$$ The first integral is simple, the second one more difficult but you arrive to $$I=\frac{\log ^2(y)}{2}-\log (y+1) \log (y)-\text{Li}_2(-y)$$
I hope this could be useful.
A: I was thinking maybe you could introduce some auxiliary integrale, with a "t" parameter in it, and which you could derive to obtain a more easily calculable expression and then get this value. 
For instance: 
The integral $I=\int_{0}^{1}\dfrac{x-1}{\ln(x)}dx$  which is a bit annoying to calculate, you have to do the same method that the one you used for your integral. But you can introduce:
$I(t)=\int_{0}^{+\infty}\dfrac{(x-1)x^t}{\ln(x)}dx$
Prove than you can differentiate under the integral, and then you can get rid of the $\ln$, which is quite convenient. See what I mean? 
