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.

$\bf{Another \; Solution::}$ Let $\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)\cdot (x\cos x-\sin x)}dx$
Now $\displaystyle I = \int\frac{2x^2}{(x^2\sin 2x+2x\cos^2 x-2x\sin^2 x-\sin 2x)}dx$
So $\displaystyle I = \int\frac{2x^2}{(x^2-1)\sin 2x+2x\cos 2x}dx$
Now $\displaystyle (x^2-1)\sin 2x+2x\cos 2x = \sqrt{(x^2-1)^2+(2x)^2}\cdot \left\{\left(\frac{x^2-1}{x^2+1}\right)\sin 2x+\left(\frac{2x}{x^2+1}\right)\cos 2x\right\}$
Now We can write $\displaystyle (x^2-1)\sin 2x+2x\cos 2x = (x^2+1)\cdot \sin (2x+\alpha)$
Where $\displaystyle \displaystyle \left(\frac{x^2-1}{x^2+1}\right) = \cos \alpha$
 and $\displaystyle \left(\frac{2x}{x^2+1}\right)=\sin \alpha$ 
and so  $\displaystyle\tan \alpha = \left(\frac{2x}{x^2-1}\right)\Rightarrow \alpha = \tan^{-1}\left(\frac{2x}{x^2-1}\right)$
So Integral $\displaystyle I = \int \csc(2x+\alpha)\cdot \left(\frac{2x^2}{x^2+1}\right)dx$
Now let $\displaystyle (2x+\alpha) = t\Rightarrow \left\{2x+\tan^{-1}\left(\frac{2x}{x^2-1}\right)\right\} = t\;,$Then $\displaystyle \left(\frac{2x^2}{x^2+1}\right)dx=dt$
So Integral $\displaystyle I = \int \csc tdt = \ln \left|\tan \frac{t}{2}\right|+\mathcal{C} = \ln \left|\tan \left(x+\frac{\alpha}{2}\right)\right|+\mathcal{C} $
