Answer
$$\frac{1}{2}\ln \left| {\frac{x}{{x + 2}}} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} + 2x}}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{1}{{{x^2} + 2x}} = \frac{1}{{x\left( {x + 2} \right)}} = \frac{A}{x} + \frac{B}{{x + 2}} \cr
& {\text{Multiplying by }}{x^2} + 2x{\text{, we have}} \cr
& {\text{1}} = A\left( {x + 2} \right) + Bx \cr
& {\text{if we set }}x = 0, \cr
& {\text{1}} = A\left( 2 \right) \cr
& A = \frac{1}{2} \cr
& {\text{if we set }}x = - 2, \cr
& {\text{1}} = A\left( 0 \right) + B\left( { - 2} \right) \cr
& B = - \frac{1}{2} \cr
& {\text{then}} \cr
& \frac{1}{{{x^2} + 2x}} = \frac{{1/2}}{x} + \frac{{ - 1/2}}{{x + 2}} \cr
& \int {\frac{{dx}}{{{x^2} + 2x}}} = \int {\left( {\frac{{1/2}}{x} + \frac{{ - 1/2}}{{x + 2}}} \right)dx} \cr
& {\text{integrating}} \cr
& = \frac{1}{2}x - \frac{1}{2}\ln \left| {x + 2} \right| + C \cr
& = \frac{1}{2}\ln \left| {\frac{x}{{x + 2}}} \right| + C \cr} $$
