How to solve the integral $\int (\frac{2}{x^3}+\frac{1}{x})\cdot \sin x\, \operatorname{d}x$

Question

I've tried a couple of substitutions, even a desperate one by parts, but nothing has worked for me.

Answer 1

Notice that

\begin{align} \int \frac{1}{x}\sin x\mathop{dx}&= -\frac{1}{x}\cos x- \int \frac{1}{x^2}\cos x\mathop{dx}\\ &=-\frac{1}{x}\cos x- \left( \frac{1}{x^2}\sin x +\int \frac{2}{x^3}\sin x\mathop{dx}\right)\\ &=-\frac{1}{x}\cos x- \frac{1}{x^2}\sin x -\int \frac{2}{x^3}\sin x\mathop{dx} \end{align} Therefore,

$$\int \left(\frac{2}{x^3}+\frac{1}{x}\right)\sin x\mathop{dx} =-\frac{1}{x}\cos x- \frac{1}{x^2}\sin x +C $$