Answer
$$ - \frac{2}{x} + 2\ln \left| {\frac{{x + 1}}{x}} \right|$$
Work Step by Step
$$\eqalign{
& \int {\frac{2}{{{x^3} + {x^2}}}dx} \cr
& {\text{integrand }} \cr
& = \frac{2}{{{x^3} + {x^2}}} \cr
& {\text{factor}} \cr
& = \frac{2}{{{x^2}\left( {x + 1} \right)}} \cr
& {\text{partial fractions}} \cr
& \frac{2}{{{x^2}\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} \cr
& 2 = Ax\left( {x + 1} \right) + B\left( {x + 1} \right) + C{x^2} \cr
& {\text{multiplying}} \cr
& 2 = A{x^2} + Ax + Bx + B + C{x^2} \cr
& 2 = \left( {A{x^2} + C{x^2}} \right) + \left( {Ax + Bx} \right) + B \cr
& {\text{by equating the coefficients}} \cr
& {x^2}:{\text{ }}A + C = 0 \cr
& x:{\text{ }}A + B = 0 \cr
& {x^0}:{\text{ }}B = 2 \cr
& {\text{Solving these equations}} \cr
& A = - 2 \cr
& B = 2 \cr
& C = 2 \cr
& {\text{substituting constants}} \cr
& \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} = \frac{{ - 2}}{x} + \frac{2}{{{x^2}}} + \frac{2}{{x + 1}} \cr
& \int {\frac{2}{{{x^3} + {x^2}}}dx} = \int {\left( {\frac{{ - 2}}{x} + \frac{2}{{{x^2}}} + \frac{2}{{x + 1}}} \right)dx} \cr
& {\text{integrating}} \cr
& = - 2\ln \left| x \right| - \frac{2}{x} + 2\ln \left| {x + 1} \right| \cr
& {\text{Simplify}} \cr
& = - \frac{2}{x} + 2\ln \left| {\frac{{x + 1}}{x}} \right| \cr} $$