The ant derivative of $\dfrac{1}{\left(1+x\tan x\right)^{2}}$ with respect to $x$ is equal to 

$\begin{aligned} I & =\int \frac{d x}{(1+x \tan x)^{2}}=\int \frac{\cos ^{2} x d x}{(x \sin x+\cos x)^{2}} \\ & =\int-\frac{\cos x}{x} d\left(\frac{1}{x \sin x+\cos {x}}\right) \\ & =-\frac{\cos x}{x} \times \frac{1}{(x \sin x+\cos x)}+\int \frac{1}{x \sin x+\cos x} d\left(\frac{\cos x}{x}\right) \\ & =-\frac{1}{x(x \tan x+1)}-\int \frac{d x}{x^{2}}= \frac{\tan x}{x \tan x+1}+C\end{aligned}$