Evaluate $\int_0^{\infty}\frac{\ln(x+\frac{1}{x})}{1+x^2}\cdot dx$ 
Evaluate: $$\int_0^{\infty}\dfrac{\ln(x+\frac{1}{x})}{1+x^2}\cdot dx$$

I tried substituting $\ln(x+\frac{1}{x})=t$ that to some extent created denominator but made it more tedious. Please give some hints to solve.
 A: Let $\displaystyle I = \int^{\infty}_{0}\ln(x+x^{-1})\cdot \frac{1}{1+x^2}dx$
Put $x=\tan \theta$ and $dx = \sec^2 \theta $ and changing limits 
So $\displaystyle I = \int^{\frac{\pi}{2}}_{0}\frac{\ln(\tan \theta+\cot \theta)}{1+\tan^2 \theta}\cdot \sec^2 \theta d \theta = \int^{\frac{\pi}{2}}_{0}\ln(\tan \theta+\cot \theta)d\theta$
$\displaystyle I  = \int^{\frac{\pi}{2}}_{0}\ln(2)-\int^{\frac{\pi}{2}}_{0}\ln (\sin2\theta)d\theta = \frac{\pi}{2}\ln(2)-\int^{\pi}_{0}\ln(\sin \theta)d\theta$
Above $\displaystyle \int^{\pi}_{0}\ln(\sin x)dx = \int^{\frac{\pi}{2}}_{0}\ln(\sin x)dx = -\pi \ln(2)$ is well known Integral
So $\displaystyle I = \frac{\pi}{2}\ln(2)+\frac{\pi}{2}\ln(2) = \pi\ln(2)$
Evaluation of $\displaystyle \int^{\frac{\pi}{2}}_{0}\ln(\sin x)dx$
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\sin u) du$$
And the substitution $v=\frac{\pi}{2}-x$ reduces the integral to,
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos v) dv$$
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos u) du$$
Now adding the integrals and noting properties of logarithms we have,
$$2I=\int_{0}^{\frac{\pi}{2}} \left( \ln (2 \sin x \cos x)-\ln 2\right) dx$$
Double angle,
$$2I=\int_{0}^{\frac{\pi}{2}} \ln (\sin 2x) dx -\frac{\pi}{2} \ln 2$$
The substitution $s=2x$ gives
$$2I=\frac{1}{2}\int_{0}^{\pi} \ln (\sin s) ds -\frac{\pi}{2} \ln 2$$
But 
$$\int_{0}^{\pi} \ln (\sin s) ds=2I$$
Follows from the substitution  $w=\frac{\pi}{2}-s$  and the evenness of the function $f(w)=\ln (\cos  w)$:
$$\int_{0}^{\pi} \ln (\sin s) ds$$
$$=-\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \ln (\cos w) dw$$
$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln (\cos w)dw $$
$$=2 \int_{0}^{\frac{\pi}{2}} \ln (\cos w) dw=2I$$
So,
$$2I=I-\frac{\pi}{2}\ln 2$$
$$I=-\frac{\pi}{2}\ln 2$$
