Answer
$$2\ln \left| x \right| + 3{\tan ^{ - 1}}\left( {x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{x^2} + 7x + 4}}{{{x^3} + 2{x^2} + 2x}}} dx \cr
& = \int {\frac{{2{x^2} + 7x + 4}}{{x\left( {{x^2} + 2x + 2} \right)}}} dx \cr
& {\text{partial fractions}} \cr
& \frac{{2{x^2} + 7x + 4}}{{x\left( {{x^2} + 2x + 2} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 2x + 2}} \cr
& {\text{multiplying}} \cr
& 2{x^2} + 7x + 4 = A\left( {{x^2} + 2x + 2} \right) + \left( {Bx + C} \right)x \cr
& 2{x^2} + 7x + 4 = A{x^2} + 2Ax + 2A + B{x^2} + Cx \cr
& 2{x^2} + 7x + 4 = \left( {A{x^2} + B{x^2}} \right) + \left( {2Ax + Cx} \right) + 2A \cr
& {\text{by equating the coefficients}} \cr
& {x^2}:{\text{ }}A + B = 2 \cr
& x:{\text{ }}2A + C = 7 \cr
& {x^0}:{\text{ }}2A = 4 \cr
& {\text{Solving these equations}} \cr
& A = 2 \cr
& B = 0 \cr
& C = 3 \cr
& {\text{substituting constants}} \cr
& \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 2x + 2}} = \frac{2}{x} + \frac{3}{{{x^2} + 2x + 2}} \cr
& \int {\frac{{2{x^2} + 7x + 4}}{{{x^3} + 2{x^2} + 2x}}} dx = \int {\left( {\frac{2}{x} + \frac{3}{{{x^2} + 2x + 2}}} \right)dx} \cr
& = \int {\frac{2}{x}dx} + \int {\frac{3}{{{x^2} + 2x + 2}}dx} \cr
& = \int {\frac{2}{x}dx} + \int {\frac{3}{{{x^2} + 2x + 1 + 1}}dx} \cr
& = \int {\frac{2}{x}dx} + \int {\frac{3}{{{{\left( {x + 1} \right)}^2} + 1}}dx} \cr
& {\text{integrating}} \cr
& = 2\ln \left| x \right| + 3{\tan ^{ - 1}}\left( {x + 1} \right) + C \cr} $$
