Answer
$$ - 2\cot x + \csc x - \ln \left| {\csc x - \cot x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2 - \cos x + \sin x}}{{{{\sin }^2}x}}} dx \cr
& {\text{Distribute the numerator}} \cr
& = \int {\left( {\frac{2}{{{{\sin }^2}x}} - \frac{{\cos x}}{{{{\sin }^2}x}} + \frac{{\sin x}}{{{{\sin }^2}x}}} \right)} dx \cr
& {\text{Simplify}} \cr
& = \int {\left( {\frac{2}{{{{\sin }^2}x}} - \left( {\frac{1}{{\sin x}}} \right)\left( {\frac{{\cos x}}{{\sin x}}} \right) + \frac{1}{{\sin x}}} \right)} dx \cr
& = 2\int {{{\csc }^2}x} dx - \int {\csc x\cot x} dx + \int {\csc } xdx \cr
& {\text{Integrate by using basic integration rules}} \cr
& = 2\left( { - \cot x} \right) - \left( { - \csc x} \right) - \ln \left| {\csc x - \cot x} \right| + C \cr
& = - 2\cot x + \csc x - \ln \left| {\csc x - \cot x} \right| + C \cr} $$
