Is there a quicker way to solve this integral: $\int \frac{3-\cos(x)}{(1+2\cos(x))\sin^2(x)}dx$?

HINT

One can look for coefficients of identity $$f(y)=\frac{3-y}{(1+2y)(1-y^2)}=\frac A{1+2y}+\frac B{1-y}+\frac C{1+y}:$$ $$A=\lim_{y\to -\dfrac12}(1+2y)f(y) =\frac{14}3,$$ $$B=\lim_{y\to 1}(1-y)f(y) = \frac13,$$ $$C=\lim_{y\to-1}(1+y)f(y) = -2$$ and then find the integrals through the universal trigonometric substitution and known integrals $$\int\dfrac{\mathrm dx}{1-\cos(x)}=\dfrac12\int\dfrac{\mathrm dx}{\sin^2\left(\dfrac x2\right)} = -\cot\left(\dfrac x2\right)+constant,$$ $$\int\dfrac{\mathrm dx}{1+\cos(x)}=\dfrac12\int\dfrac{\mathrm dx}{\cos^2\left(\dfrac x2\right)} = \tan\left(\dfrac x2\right)+constant.$$