integral of {log(x +1/x)} .{1/1+x^2}dx (limit is from 0 to infinity ) is

$$\begin{gathered} {\int }_0^{\infty }\left\{\log \left(x+\frac{1}{x}\right)\right\}\frac{1}{1+x^2}dx \\ Put x = \tan t or dx = se{c}^{2}t dt and limit changes from 0 to \pi /2 \\ \log \left(x+\frac{1}{x}\right)=\log \left(\frac{1+{\tan }^{2}t}{\tan t}\right)=\log \left(\frac{1}{\sin t.\cos t}\right)=-\log \sin t -\log \cos t \\ \frac{1}{1+x^2}=\frac{1}{se{c}^{2}t} \\ Putting all these value we get, \\ -{\int }_0^{\pi /2}\left(\log \sin t +\log \cos t\right)dt \\ {\int }_0^{\pi /2}\log \sin t dt={\int }_0^{\pi /2}\log \sin \left(\frac{\pi }{2}-t \right)dt={\int }_0^{\pi /2}\log \cos t dt \\ -{\int }_0^{\pi /2}\left(\log \sin t +\log \cos t\right)dt=-2{\int }_0^{\pi /2}\log \sin tdt \end{gathered}$$

Now we can write it as, 
$$\begin{gathered} -2{\int }_0^{\pi /2}\log \left(\sin x\right)dx \\ Now finding {\int }_0^{\pi /2}\log \left(\sin x\right)dx, \end{gathered}$$
Let I=${\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin x\right)dx$ -------(1)
$\Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left[\sin \left(\frac{\mathrm{\pi }}{2}-x\right)\right]dx \left[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\right]$
$\Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\cos x\right)dx$  --------(2)

Eq(1)+Eq(2)$\Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin x\right)+{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\cos x\right)$

$$\begin{gathered} \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\log \left(\sin x\right)+\log \left(\cos x\right)\right]dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin x \cos x\right)dx \end{gathered}$$

$$\begin{gathered} \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{\sin 2x}{2}\right)dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin 2x\right)dx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2 dx -------\left(3\right) \end{gathered}$$

Let 2x=t
$\Rightarrow dx=\frac{dt}{2}-----\left(4\right)$

When x=0, t=0.
When x=$\frac{\mathrm{\pi }}{2}, t=\mathrm{\pi }$

So, using Eq(4) in Eq(3), we have:-

$$\begin{gathered} 2I={\int }_0^{\mathrm{\pi }}\log \left(\sin t\right)\frac{dt}{2}-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2 dx \\ \Rightarrow 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin t\right)dt-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2 dx \end{gathered}$$

$$\begin{gathered} \Rightarrow 2I=I-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2 dx \\ \Rightarrow I=-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2 dx \end{gathered}$$

$$\begin{gathered} \Rightarrow I=\log 2{\left[-x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=-\frac{\mathrm{\pi }}{2}\log 2 \end{gathered}$$
So
$$\begin{gathered} -2I = -2\times \left(-\frac{\pi }{2}\log 2\right) \\ =\pi \log 2 \end{gathered}$$