Question

Solve
${\int }_0^{\infty }\ell n(x+\frac{1}{x}).\frac{dx}{1+x^2}$

Answer 1

$\begin{array}{c}{\int }_0^{\infty }\log \left(x+\frac{1}{x}\right)\cdot \frac{dx}{\left(1+x^2\right)}\\ Now,\\ puttingx=\tan \theta \\ \theta ={\tan }^{-1}x\\ ={\int }_0^{\frac{\pi }{2}}\frac{\log \left(\tan \theta +\cot \theta \right)}{(1+{\tan }^2\theta )}\cdot {\sec }^2\theta d\theta \\ ={\int }_0^{\frac{\pi }{2}}\frac{\log \left|\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cdot \cos \theta }\right|}{{\sec }^2\theta }\cdot {\sec }^2\theta d\theta \\ ={\int }_0^{\frac{\pi }{2}}\log \left(\frac{1}{\sin \theta \cdot \cos \theta }\right)d\theta \\ =-{\int }_0^{\frac{\pi }{2}}\left\{\log \left(\sin \theta \right)+\log \left(\cos \theta \right)\right\}d\theta \\ =\frac{\pi }{2}\log 4\end{array}$

Hence,solved.