Answer
$$2\ln \left| {x + 1} \right| + \frac{1}{{x + 1}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{x^2} + 3x + 3}}{{{{\left( {x + 1} \right)}^3}}}} dx \cr
& {\text{The partial fraction decomposition of the integrand is}} \cr
& \frac{{2{x^2} + 3x + 3}}{{{{\left( {x + 1} \right)}^3}}} = \frac{A}{{x + 1}} + \frac{B}{{{{\left( {x + 1} \right)}^2}}} + \frac{C}{{{{\left( {x + 1} \right)}^3}}} \cr
& {\text{Multiplying by }}{\left( {x + 1} \right)^3}{\text{ yields}} \cr
& 2{x^2} + 3x + 3 = A{\left( {x + 1} \right)^2} + B\left( {x + 1} \right) + C \cr
& 2{x^2} + 3x + 3 = A\left( {{x^2} + 2x + 1} \right)Bx + B + C \cr
& 2{x^2} + 3x + 3 = A{x^2} + 2Ax + A + Bx + B + C \cr
& {\text{Collecting like powers of }}x{\text{, this becomes}} \cr
& 2{x^2} + 3x + 3 = A{x^2} + \left( {2Ax + Bx} \right) + A + B + C \cr
& {\text{Equating corresponding coefficients yields the following system }} \cr
& {\text{of linear equations}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,A = 2 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,2A + B = 3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,A + B + C = 3 \cr
& {\text{Solving the system of linear equations we obtain}} \cr
& \,\,\,A = 2,\,\,\,B = - 1,\,\,\,C = 2 \cr
& \cr
& {\text{Then}}{\text{, the integrand can be written as}} \cr
& \frac{{2{x^2} + 3x + 3}}{{{{\left( {x + 1} \right)}^3}}} = \frac{2}{{x + 1}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}} + \frac{2}{{{{\left( {x + 1} \right)}^3}}} \cr
& \int {\frac{{2{x^2} + 3x + 3}}{{{{\left( {x + 1} \right)}^3}}}} dx = \int {\left( {\frac{2}{{x + 1}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}} + \frac{2}{{{{\left( {x + 1} \right)}^3}}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = 2\ln \left| {x + 1} \right| + \frac{1}{{x + 1}} + 2\left( {\frac{{{{\left( {x + 1} \right)}^{ - 2}}}}{{ - 2}}} \right) + C \cr
& = 2\ln \left| {x + 1} \right| + \frac{1}{{x + 1}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}} + C \cr} $$
