How to integrate this indefinite integral?

Question

$$\int\frac{x^2\sec^2x}{(x \tan{x}+1)^{2}}\,\mathrm{d}x$$

I tried the online available calculators but they cannot calculate the answer or provide the solution.

Answer 1

This is basically working backwards, like @Arthur suggests:

\begin{align} \frac{x^2\sec^2 x}{(x\tan x+1)^2} &= \frac{x^2}{\left(x\frac{\sin x}{\cos x} + 1\right)^2\cos^2 x}\\ &= \frac{x^2}{(x\sin x + \cos x)^2}\\ &= \frac{x^2(\sin^2 x + \cos ^2 x) - x\cos x\sin x + x \cos x\sin x}{(x\sin x + \cos x)^2}\\ &= \frac{x\sin x(x \sin x + \cos x) - (\sin x - x\cos x)x\cos x}{(x\sin x + \cos x)^2}\\ &= \frac{(\sin x - x\cos x)'(x \sin x + \cos x) - (\sin x - x\cos x)(x\sin x + \cos x)'}{(x\sin x + \cos x)^2}\\ &= \left(\frac{\sin x - x\cos x}{x \sin x + \cos x}\right)' \end{align}

so $$\int \frac{x^2\sec^2 x}{(x\tan x+1)^2} \,dx = \frac{\sin x - x\cos x}{x \sin x + \cos x}$$