integration - find $I=int_0^{infty} log{(x+1/x)},frac{dx}{1+x^2}$

Using $$\int_0^{\pi/2} \log\sin x\,\mathrm dx= -\dfrac{\pi}{2} \log 2$$ how to find

$$I=\int_0^{\infty} \log{(x+1/x)}\,\frac{dx}{1+x^2}.$$

Putting $x=\tan z$,

$I=\int_0^{\pi/2} (\log 2-\log(\sin(2z)))dz=\frac{\pi}{2}\log 2-1/2\int_0^{\pi} \log(\sin(u))du$ for $2z=u$

what to do next?

Answer

Split the second integral:

$$\int_0^{\pi}\log(\sin(u)) du = \int_0^{\pi/2}\log(\sin(u)) du + \int_{\pi/2}^{\pi}\log(\sin(u)) du$$

And use a change of variable $u=\pi-x$ for the second part.