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: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$

Here I give another way to solve the problem. We rewrite the integral as
\begin{eqnarray}
I&=&\int_0^\infty\log(x+\frac{1}{x})\frac{1}{1+x^2}dx\\
&=&\int_0^\infty\frac{\log(1+x^2)}{1+x^2}dx-\int_0^\infty\frac{\log x}{1+x^2}dx
\end{eqnarray}
Note
$$ \int_0^\infty\frac{\log(1+x^2)}{1+x^2}dx=\pi\log 2$$
from Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$
and
$$ \int_0^\infty\frac{\log x}{1+x^2}dx=0. $$
Thus we have
$$ I=\pi \log 2.$$ 
