Integration of trigonometric function $\int\frac{\sin(2x)}{\sin(x)-\cos(x)}dx$ 
$$\int\frac{\sin(2x)}{\sin(x)-\cos(x)}dx$$

My attempt: Firstly, $\sin(2x)=2\sin(x)\cos(x)$.
After that, eliminate the $\cos(x)$ seen in both the numerator and denominator to get
$$2\int\frac{\sin(x)}{\tan(x)-1}\ dx.$$
From here onwards, should I convert $\sin(x)$, $\tan(x)$ to half-angles and use $\tan(x/2)=t$?
But this would be a time consuming method. Any suggestions?

HINT:
$$\int\frac{\sin(2x)}{\sin(x)-\cos(x)}\space\text{d}x=$$

Use $\sin(2x)=2\sin(x)\cos(x)$:

$$2\int\frac{\sin(x)\cos(x)}{\sin(x)-\cos(x)}\space\text{d}x=$$

Sustitute $u=\tan\left(\frac{x}{2}\right)$ and $\text{d}u=\frac{x\sec^2\left(\frac{x}{2}\right)}{2}\space\text{d}x$:

$$-8\int\frac{u(u^2-1)}{(u^2+1)^2(u^2+2u-1)}\space\text{d}u$$
Now, use partial fractions.
