Different way $\int_{0}^{\infty}e^{-2x}\frac{\ln \cosh(x)}{1+\cosh(x)}dx$ I would like to evaluate this integral, $$I=\int_{0}^{\infty}e^{-2x}\frac{\ln \cosh(x)}{1+\cosh(x)}\mathrm dx$$
An approach might would be:


*

*Remove the $\cosh(x)$


$$I=2\int_{0}^{\infty}e^{-x}\ln\left(\frac{e^x+e^{-x}}{2}\right)\frac{dx}{(e^x+1)^2}$$


*Make a substitution of: $u=e^x$, $du=e^x dx$


$$I=2\int_{1}^{\infty}\frac{\ln(1+u^2)}{(u+u^2)^2}du-2\int_{1}^{\infty}\frac{\ln(2u)}{(u+u^2)^2}du=2(I_1-I_2)$$


*Partial fraction decomp of integral $I_1$ reveals


$$I_1=2\int_{1}^{\infty}\frac{\ln(1+x^2)}{1+x}dx+\int_{1}^{\infty}\frac{\ln(1+x^2)}{(1+x)^2}dx-2\int_{1}^{\infty}\frac{\ln(1+x^2)}{x}dx+\int_{1}^{\infty}\frac{\ln(1+x^2)}{x^2}dx$$
$$I_2=2\int_{1}^{\infty}\frac{\ln(2x)}{1+x}dx+\int_{1}^{\infty}\frac{\ln(2x)}{(1+x)^2}dx-2\int_{1}^{\infty}\frac{\ln(2x)}{x}dx+\int_{1}^{\infty}\frac{\ln(2x)}{x^2}dx$$
abandomed this approach, because too much work

I was thinking of applying integration by parts, $u=\ln\cosh(x)$, $du=\tanh(x) dx$
$dv=\frac{e^{-2x}}{1+\cosh(x)}dx=\frac{2^{-x}}{(e^x+1)^2}dx$
but, when I try to evaluate, it is not possible.
Making a sub: $u=e^x$ and partial fraction decomposition, it is can easily integrate to $$v=4\ln(1+e^x)-\frac{2}{1+e^x}-2e^{-x}-4x$$

This is getting very bored, unfortunately I don't know any other methods. Can anybody evaluate integral $I$ by another approach? 
 A: Let $I$ denote the integral. Then applying the substitution $u = e^{-x}$ followed by integration by parts,
\begin{align*}
I
&= \int_{0}^{1} \frac{2u^2}{(1+u)^2} \log\left(\frac{1+u^2}{2u}\right)\,du \\
&= \int_{0}^{1} 2\left( 1+u - \frac{1}{1+u} - 2\log(1+u)\right)\frac{1-u^2}{u(1+u^2)} \, du \\
&= \underbrace{ \int_{0}^{1} \left( \frac{6}{1+u^2} -\frac{2u}{1+u^2} -2 \right) \, du}_{=(1)} -4 \underbrace{ \int_{0}^{1} \frac{1-u^2}{u(1+u^2)} \log(1+u) \, du }_{=(2)}
\end{align*}
Now $(1)$ is easily computed by direct computation, yielding $(1) = \frac{3\pi}{2} - \log 2 - 2$. For $(2)$, we utilize the identity $\log(1+u)=\int_{0}^{1} \frac{u}{1+uv} \,dv$ to write 
$$ (1) = \int_{0}^{1} \int_{0}^{1} \frac{1-u^2}{(1+vu)(1+u^2)} \, dudv. $$
Interchanging the role of $u$ and $v$ does not alter the value of integral, thus we find that
\begin{align*}
(1)
&= \frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{1}{1+vu}\left( \frac{1-u^2}{1+u^2} + \frac{1-v^2}{1+v^2} \right) \, dudv \\
&= \int_{0}^{1} \int_{0}^{1} \frac{1-uv}{(1+u^2)(1+v^2)} \, dudv
= \left(\frac{\pi}{4}\right)^2 - \left( \frac{\log 2}{2}\right)^2.
\end{align*}
(The last step easily follows since each term of the double integral factors into the product of two 1-dimensional integrals.) Therefore we have

$$ I = \log^2 2 - \log 2 - \frac{\pi^2}{4} + \frac{3\pi}{2} - 2.$$

A: As  LittleCuteKemono commented, at step $3$, there are problems.
What I did was to use step $(2)$. The antiderivatives can be computed but the problem is that the result involves  polylogarithm functions. However, the results can be obtained as
$$I_1=\frac{3 \pi }{4}-\frac{7 \pi ^2}{24}-\frac{3 \log ^2(2)}{2}+2\log (2)$$
$$I_2=1-\frac{\pi ^2}{6}-2 \log ^2(2)+\frac{5 \log (2)}{2}$$ making
$$I=\frac{1}{4} (2 +2\log (2)-\pi) (\pi +2\log (2)-4)$$ identical to the result given by Sangchul Lee in his/her elegant answer (it appeared when I was finising typing this).
