Now this is one of the ones that I am most proud of

$\displaystyle \int_0^{\infty}\ln\left(\frac{e^x+1}{e^x-1}\right)~dx$

First let

$\displaystyle \ln(u)=x$

$\displaystyle \Rightarrow{\frac{du}{u}=dx}$

And we can see that

$\displaystyle 0\mapsto{1}$

$\displaystyle \infty\mapsto\infty$

So we get

$\displaystyle =\int_1^{\infty}\frac{\ln\left(\frac{u+1}{u-1}\right)}{u}~dx$

Now here is the next step

Let

$\displaystyle u=\frac{1}{z}$

$\displaystyle \Rightarrow{du=\frac{-dz}{z^2}}$

$\displaystyle \infty\mapsto{0}$

$\displaystyle 1\mapsto{1}$

$\displaystyle =-\int_{1}^{0}\frac{\ln\left(\frac{\frac{1}{z}+1}{\f rac{1}{z}-1}\right)}{\frac{1}{z}}\cdot\frac{dz}{z^2}$

Now with a little mutliplicative identity

$\displaystyle =\int_0^1\frac{\ln\left(\frac{1+z}{1-z}\right)}{z}~dz$

$\displaystyle =\int_0^1\frac{\ln(1+z)}{z}~dz-\int_0^{1}\frac{\ln(1-z)}{z}~dz$

Now imputting the uniformly convergent power seires

$\displaystyle \ln(1+z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{n+1}}{n+1}~\wedge~\ln(1-z)=-\sum_{n=0}^{\infty}\frac{z^{n+1}}{n+1}$

We ultimately arrive at (I've doen these earlier in the thread if you are curious)

$\displaystyle =\frac{\pi^2}{12}+\frac{\pi^2}{6}$

$\displaystyle =\frac{\pi^2}{4}$

So now industrious as ever we have found that

$\displaystyle \boxed{\int_0^{\infty}\ln\left(\frac{e^x+1}{e^x-1}\right)~dx=\int_1^{\infty}\frac{\ln\left(\frac{x +1}{x-1}\right)}{x}~dx=\int_0^1\frac{\ln\left(\frac{1+x} {1-x}\right)}{dx}~dx=\frac{\pi^2}{4}}\quad\blacksquar e$