How to calculate the improper integral $\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx$ How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$

Put $x = \tan{\theta}$. Then you have 
\begin{align*}
I &= \int_{0}^{\pi/2} \log(\tan\theta + \cot\theta) \ d\theta \\\ &= \int_{0}^{\pi/2} \log\frac{2}{\sin{2\theta}} \ d\theta \\\ &= - \int_{0}^{\pi/2} (\log{\sin\theta} +\log{\cos\theta}) d\theta \\\ &= -2 \int\limits_{0}^{\pi/2} \log\: {\sin\theta} \ d\theta = \pi \: \log{2}
\end{align*}
