Is there a way of solving integrals where the numerator is an integral of the denominator? Is there a way of solving integrals where the numerator is an integral of the denominator?

I was evaluating the integral $$\int \frac{x-\sin x}{1-\cos x}\mathrm{d}x$$. I separated the numerator into two fractions and solved it.

However, the numerator seems to be the integral of the denominator. So, is there an alter method to do this integral by taking advantage of this fact?
$\int \frac{x-\sin x}{1-\cos x}\mathrm dx$ 
$\int\frac x{1-\cos x}dx-\int\frac{\sin x}{1-\cos x} dx$ 
Integrate by parts for the first one and then substitute $t=\tan(\frac{x}2)$  
For the second one simple put $\cos x=u$ 
As far as I know, for $\int\frac{f(x)}{f'(x)}dx$, I don't think there is a unique method.
$\int \frac{x-\sin x}{1-\cos x}\mathrm dx$ 
$\int\frac x{1-\cos x}dx-\int\frac{\sin x}{1-\cos x} dx$ 
Integrate by parts for the first one and then substitute $t=\tan(\frac{x}2)$  
For the second one simple put $\cos x=u$ 
As far as I know, for $\int\frac{f(x)}{f'(x)}dx$, I don't think there is a unique method.