${\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{{\left(x\tan x+1\right)}^2}=}$

${\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{{\left(x\tan x+1\right)}^2}=}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{{\cos }^{2}xdx}{{\left(x\sin x+\cos x\right)}^2}}$

                             $={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\cos x}{x}\cdot \frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx}$

                            $={\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\cos x}{x}\cdot \frac{d}{dx}\left(\frac{-1}{\left(x\sin x+\cos x\right)}\right)dx}$

                             $={\left[\frac{-\cos x}{x\left(x\sin x+\cos x\right)}+\frac{1}{x}\right]|}_0^{\frac{\pi }{4}}={\frac{\tan x}{x\tan x+1}|}_0^{\frac{\pi }{4}}$

                             $=\frac{4}{4+\pi }$