Is there a quicker way to solve this integral: $\int \frac{3-\cos(x)}{(1+2\cos(x))\sin^2(x)}dx$? The integral is:
$$ \int \frac{3-\cos(x)}{(1+2\cos(x))\sin^2(x)}dx$$
This is the way I approached it:
$$ \tan\left(\frac{x}{2}\right)=u\\dx=\frac{2}{\sec^2\left(\frac{x}{2}\right)}du$$
By using trigonometric identities we get:
$$ \sin(x)=\frac{2u}{1+u^2};\ \cos(x)=\frac{1-u^2}{1+u^2};\ \sec^2\left(\frac{x}{2}\right)=1+u^2
$$
Therefore the integral now becomes:
$$ 2\int \frac{3-\frac{1-u^2}{1+u^2}}{\left(1+2\left(\frac{1-u^2}{1+u^2}\right)\right)\left(\frac{2u}{1+u^2}\right)^2(1+u^2)}du=$$ $$\int\frac{(1+2u^2)(1+u^2)}{u^2(3-u^2)}du$$ By dividing the two polynomials we get: $$\int\left(-2-\frac{9u^2+1}{u^2(3-u^2)}\right)du$$ Using partial fractions we get to the simplified form: $$\int\left(-2-\frac{1}{3u^2}+\frac{28}{3(3-u^2)}\right)du$$ $$-2u-\frac{1}{3u}+\frac{28\sqrt3}{9}\int\frac{1}{1-\left(\frac{u}{\sqrt3}\right)^2}du \\ -2u-\frac{1}{3u}+\frac{28\sqrt3\tanh^{-1}{\left(\frac{u}{\sqrt3}\right)}}{9}+C$$
By substituting back in for $u$, we get the solution: $$ \bbox[5px,border:2px solid black]{\frac{28\sqrt3\tanh^{-1}{\left(\frac{\tan\left(\frac{x}{2}\right)}{\sqrt3}\right)}}{9}-2\tan\left(\frac{x}{2}\right)-\frac{1}{3}\cot\left(\frac{x}{2}\right)+C}$$
My question is, as you can understand from the title, is there any easier and faster way to solve this integral? If so, how? Thank you.

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
$$\int\dfrac{\mathrm dx}{1-\cos(x)}=\dfrac12\int\dfrac{\mathrm dx}{\sin^2\left(\dfrac x2\right)} = -\cot\left(\dfrac x2\right)+constant,$$
$$\int\dfrac{\mathrm dx}{1+\cos(x)}=\dfrac12\int\dfrac{\mathrm dx}{\cos^2\left(\dfrac x2\right)} = \tan\left(\dfrac x2\right)+constant.$$
