Answer
$$\ln \left| {x + 2} \right| + \frac{4}{{\left( {x + 2} \right)}} + \frac{2}{{{{\left( {x + 2} \right)}^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{{{\left( {x + 2} \right)}^3}}}} dx \cr
& {\text{The partial fraction decomposition of the integrand is}} \cr
& \frac{{{x^2}}}{{{{\left( {x + 2} \right)}^3}}} = \frac{A}{{x + 2}} + \frac{B}{{{{\left( {x + 2} \right)}^2}}} + \frac{C}{{{{\left( {x + 2} \right)}^3}}} \cr
& {\text{Multiplying by }}{\left( {x + 2} \right)^3}{\text{ yields}} \cr
& {x^2} = A{\left( {x + 2} \right)^2} + B\left( {x + 2} \right) + C \cr
& {x^2} = A\left( {{x^2} + 4x + 4} \right) + B\left( {x + 2} \right) + C \cr
& {x^2} = A{x^2} + 4Ax + 4A + Bx + 2B + C \cr
& {\text{Collecting like powers of }}x{\text{, it becomes}} \cr
& {x^2} = A{x^2} + \left( {4Ax + Bx} \right) + 4A + 2B + C \cr
& {\text{Equating corresponding coefficients yields the following system }} \cr
& {\text{of linear equations}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A = 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4A + B = 0 \cr
& \,4A + 2B + C = 0 \cr
& {\text{Solving the system of linear equations we obtain}} \cr
& \,\,\,A = 1,\,\,\,B = - 4,\,\,\,C = 4 \cr
& {\text{Then}}{\text{, the integrand can be written as}} \cr
& \frac{{{x^2}}}{{{{\left( {x + 2} \right)}^3}}} = \frac{1}{{x + 2}} - \frac{4}{{{{\left( {x + 2} \right)}^2}}} + \frac{4}{{{{\left( {x + 2} \right)}^3}}} \cr
& \int {\frac{{dx}}{{x\left( {{x^2} + 1} \right)}}} = \int {\left( {\frac{1}{{x + 2}} - \frac{4}{{{{\left( {x + 2} \right)}^2}}} + \frac{4}{{{{\left( {x + 2} \right)}^3}}}} \right)} dx \cr
& = \int {\frac{1}{{x + 2}}} dx - 4\int {{{\left( {x + 2} \right)}^{ - 2}}} dx + 4\int {{{\left( {x + 2} \right)}^{ - 3}}} dx \cr
& {\text{Integrating}} \cr
& = \ln \left| {x + 2} \right| - 4\left[ {\frac{{{{\left( {x + 2} \right)}^{ - 1}}}}{{ - 1}}} \right] + 4\left[ {\frac{{{{\left( {x + 2} \right)}^{ - 2}}}}{{ - 2}}} \right] + C \cr
& = \ln \left| {x + 2} \right| + \frac{4}{{\left( {x + 2} \right)}} + \frac{2}{{{{\left( {x + 2} \right)}^2}}} + C \cr} $$
