How to solve ∫0∞log⁡(x+1x)1+x2dx ?

You may write
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\log (x+\frac{1}{x})}{1+x^2}dx={\int }_0^{\mathrm{\infty }}\frac{\log (1+x^2)-\log x}{1+x^2}dx}$
${\displaystyle ={\int }_0^{\mathrm{\infty }}\frac{\log (1+x^2)}{1+x^2}dx-{\int }_0^{\mathrm{\infty }}\frac{\log x}{1+x^2}dx}$
Clearly, by the change of variable $\to \frac{1}{x}$, we get
${\displaystyle ={\int }_0^{\mathrm{\infty }}\frac{\log x}{1+x^2}dx=-{\int }_0^{\mathrm{\infty }}\frac{\log x}{1+x^2}dx}$
On the other hand, by the change of variable
${\displaystyle x=\tan \theta ,dx=(1-{\tan }^2\theta )d\theta ,1+{\tan }^2\theta =\frac{1}{{\cos }^2\theta }}$
we obtain the classic evaluation:
${\displaystyle ={\int }_0^{\mathrm{\infty }}\frac{\log (1+x^2)}{1+x^2}dx=-2{\int }_0^{\frac{\pi }{2}}\log \left(\cos \theta \right)d\theta =\pi \log 2}$

Instead of using the trigonometric subsitution you may use the identity
${\displaystyle \log (1+x^2)={\int }_0^{1}da\frac{x^2}{1+a{x}^2}}$
Calling your integral of interest ${\displaystyle I}$ It follows that
${\displaystyle I={\int }_0^{1}da\underset{\text{use Residue Theorem to calculate this integral}}{\underbrace{{\int }_0^{\mathrm{\infty }}dx\frac{x^2}{(x^2+1)(1+a{x}^2)}}}}$
${\displaystyle =\underset{\text{use }a=b^2\text{ to turn this into a standard integra}}{\underbrace{{\int }_0^{1}da\frac{\pi }{2(a+\sqrt{a})}}}=\pi \log \left(2\right)}$
In agreement with Olivier’s answer

$$\begin{gathered} x\to \tan x \\ I=-{\int }_0^{\frac{\pi }{2}}\log (\sin x\cos x)dx \\ =-{\int }_0^{\frac{pi}{2}}\log (\sin x)dx+{\int }_0^{\frac{\pi }{2}}\log (\cos x)dx \\ =-2{\int }_0^{\frac{\pi }{2}}\log (\cos x)dx \\ =\pi \log 2 \end{gathered}$$