Integrate : $\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}dx$ 
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$

My approach : 
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$
But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.

HINT :
Rewrite the integrand
$$
\frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}
$$
as
$$
\frac{x\cos x}{x\sin x +\cos x}+\frac{x\sin x}{x\cos x -\sin x}
$$
then
$$
\frac{\color{red}{\sin x}+x\cos x-\color{red}{\sin x}}{x\sin x +\cos x}+\frac{\color{blue}{\cos x}+x\sin x-\color{blue}{\cos x}}{x\cos x -\sin x}.
$$
Now let $u=x\sin x +\cos x$ and $v=x\cos x -\sin x$.
