About integral of $\ln(e^x+1)$

Question

I'm trying to evaluate the area of the region enclosed by $y=0,\ y=x,\ and \ y=\ln(e^x+1)$.
I separate it into these two integrals: $$I=\int_{-\infty}^0{\ln(e^x+1)dx}\ and\ J=\int_0^\infty{\ln(e^x+1)-xdx}$$ I mentioned: $$\int{\ln(e^x-1)dx}=-\operatorname{Li_2}(-e^x)+C$$ So for the first integral: $$I=-\operatorname{Li_2}(-1)+\operatorname{Li_2}(0)\\=\frac{\pi^2}{12}$$ For the second integral: $$J=\lim_{x\to+\infty}(-\operatorname{Li_2}(-e^x)-\frac{x^2}2-(-\frac{\pi^2}{12}))$$ And I'm stuck here.
Also, I think using this method is "not beautiful". I tried finding another one but failed.

Answer 1

Hint: $$J = \int_{0}^{\infty}\ln(1+e^x)-\ln(e^x) \, dx = \int_{0}^{\infty}\ln(1+e^{-x})dx = \int_{-\infty}^{0}\ln(1+e^{y})dy.$$