TITLE: Evaluate $\int \limits _{0}^{\infty}\ln\left({x+\frac{1}{x}}\right)\cdot\frac{\mathrm dx}{1+x^2}$
QUESTION [5 upvotes]: $$\int \limits _{0}^{\infty}\ln\left({x+\frac{1}{x}}\right)\cdot\frac{dx}{1+x^2}$$
we are asked to solve this definite integral so here's what i did
$$\int \limits_{0}^{\infty}\ln \left({\frac{x^2 +1}{x}}\right)\cdot\frac{dx}{1+x^2} = \int\limits_{0}^{\infty}\ln \left(x^2+1\right)\cdot\frac{dx}{1+x^2} - \int \limits_{0}^{\infty}\ln (x).\frac{dx}{1+x^2}$$ now how to proceed after that ? should i intregrate both seperate functions by substituting $(x^2+1)$ and what should i susbtitute in other integral , by-parts integration is making troubles when substituting $\infty$ , now what to do ?

REPLY [3 votes]: By symmetry the integral is $$2\int_1^{\infty}\frac{\ln(x+1/x)}{x+1/x}\frac{dx}{x}$$
Make the substitution $$x=e^{\cosh^{-1}\csc t}$$ to get $$\frac{\pi}{2}\ln 2-\int_0^{\pi/2}\ln(\sin t)dt=\pi\ln 2$$