I tried possible way of getting the integral of the following integrand:
$$\int_0^{\infty} \frac{\ln\left(x+\frac{1}{x}\right)}{1+x^2}~dx$$
Firstly, I tried to write $\ln\left(x+\frac{1}{x}\right)$ as $\ln(x^2+1)-\ln(x)$ and tried to apply biparts but couldn't take it to its destination solution.
Secondly, I tried to substitute $t=x+\frac{1}{x}$ and proceeded, but couldn't solve further.
Nothing seems to work. This was a question from my exam. I don't understand how I'm supposed to solve this question in a short time when I can't solve it at home. I'm guessing there has to be a short but smart solution that is pretty much not striking me.
First step is put $x=\tan (t) $ that substitution gets rid of the denominator . Then after that we have the integral as $\int _0 ^{\frac {\pi}{2}} \log ( 1+\tan^2 (t))-\log(tan (t)) $ now use $1+tan^2(t)=sec^2 (t)$ . Now as in $0-\pi/2$ sec is positive we can write $\ln (\sec^2 (t))=2\ln (\sec (t))$ . Thus the integral is $\int _0 ^{\frac{\pi}{2}} log (\frac{\sec(t)}{\tan (t)})=\int _0 ^{\frac {\pi}{2}}-\log(\sin (t)) $ . Now I am sure you can handle this integral. The result should be $\frac {\pi}{}\log (2) $