How to solve $\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$?

You may write $$ \begin{align} \int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx&=\int_0^{\infty} \frac{\log(1+x^2)-\log x}{1+x^2}dx\\\\ &=\int_0^{\infty} \frac{\log(1+x^2)}{1+x^2}dx-\int_0^{\infty} \frac{\log x}{1+x^2}dx. \end{align} $$ Clearly, by the change of variable $ x \to \dfrac 1 x$, we get $$ \int_0^{\infty} \frac{\log x}{1+x^2}dx=-\int_0^{\infty} \frac{\log x}{1+x^2}dx=0. $$ On the other hand, by the change of variable $$x= \tan \theta, \quad dx =(1+\tan^2 \theta)d\theta, \quad 1+ \tan^2 \theta=\dfrac{1}{\cos^2 \theta},$$ we obtain the classic evaluation: $$ \int_0^{\infty} \frac{\log(1+x^2)}{1+x^2}dx=-2\int_0^{\pi/2} \log ( \cos \theta) \: d\theta=\pi \log 2. $$

Instead of using the trigonometric subsitution you may use the identity

$$\log(1+x^2)=\int_{0}^1 da \frac{x^2}{1+ax^2}$$

Calling your integral of interest $I$ It follows that

$$ I=\int_{0}^1 da \underbrace{\int_{0}^{\infty}dx\frac{x^2}{(x^2+1)(1+ax^2)}}_{\text{use Residue Theorem to calculate this integral}}\\ = \underbrace{\int_{0}^1 da\frac{\pi }{2 \left(a+\sqrt{a}\right)}}_{\text{use}\quad a=b^2\quad \text{to turn this into a standard integral}}=\pi \log (2) $$

In agreement with Olivier’s answer

Approach 1: \begin{align} I&=\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx\\ &=\int_0^{1} \frac{\log(x+\frac{1}{x})}{1+x^2}dx+\int_1^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx\\ &=2\int_0^{1} \frac{\log(x+\frac{1}{x})}{1+x^2}dx\\ &=2\int_2^{\infty} \frac{\log(u)}{u\sqrt{u^2-4}}du\\ \end{align} using $u=x+\frac{1}{x}$ in the last step. Now set $u=2\sec y$ to obtain

\begin{align} I&=\int_0^{\pi/2} \log(2\sec y)dy\\ &=\frac{\pi}{2}\log2-\int_0^{\pi/2} \log(\cos y)dy\\ &=\pi\log2 \end{align}

using this in the last step.

Approach 2:

$x \to \tan x$ \begin{align} I&=-\int_0^{\pi/2} \log(\sin x\cos x)dx\\ &=-\int_0^{\pi/2} \log(\sin x)dx+\int_0^{\pi/2} \log(\cos x)dx\\ &=-2\int_0^{\pi/2} \log(\cos x)dx\\ &=\pi \log 2 \end{align}