Guidelines for solving $\int \frac{f(x)}{f'(x)} \:dx$

Question

I'm aware of $\int \frac{f'(x)}{f(x)} \:dx$ as a model solution
The question that prompted this thought was : $$\int \frac{x + \sin(x)}{1 + \cos(x)} dx$$
I do know how to solve it but I find the trigonometric solution a bit ick-y and was wondering if a model, perhaps even a guideline could help me face questions in the format given in the title, better.

Any help is appreciated.

Answer 1

As Tyma Gaidash suggested, $$ \begin{aligned} \quad & \int \frac{(x+\sin x)(1-\cos x)}{1-\cos ^2 x} d x \\= & -\int(x+\sin x-x \cos x-\sin x \cos x) d(\cot x) \\ = & -(x+\sin x)(1-\cos x) \cot x+\int \cot x(1+\cos x+x \sin x-\cos x-\cos 2 x) d x \\ = & -(x+\sin x)(1-\cos x) \cot x+ \int \cot x\left(2 \sin ^2 x+x \sin x\right) d x =… \end{aligned} $$ Wish it helps.

Answer 2

In terms of $y = \frac{x}{2}$, the integral is $$2 \int (1 \cdot \tan y + y \cdot \sec^2 y)\,dy = 2 \int (y \tan y)' \,dy = 2 y \tan y + C = x \tan \frac{x}{2} + C. $$