Answer
$$\frac{1}{{x + 1}} + \ln \left| {\left( {x + 1} \right)\left( {{x^2} + 4} \right)} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{3{x^3} + 4{x^2} + 6x}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 4} \right)}}} dx \cr
& {\text{Decompose }}\frac{{3{x^3} + 4{x^2} + 6x}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 4} \right)}}{\text{ into partial fractions}} \cr
& \frac{{3{x^3} + 4{x^2} + 6x}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 4} \right)}} = \frac{A}{{x + 1}} + \frac{B}{{{{\left( {x + 1} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 4}} \cr
& {\text{Multiply both sides by }}{\left( {x + 1} \right)^2}\left( {{x^2} + 4} \right) \cr
& 3{x^3} + 4{x^2} + 6x = A\left( {x + 1} \right)\left( {{x^2} + 4} \right) + B\left( {{x^2} + 4} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {Cx + D} \right){\left( {x + 1} \right)^2} \cr
& {\text{Multiply}} \cr
& 3{x^3} + 4{x^2} + 6x = A{x^3} + A{x^2} + 4Ax + 4A + B{x^2} + 4B + C{x^3} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2C{x^2} + Cx + D{x^2} + 2Dx + D \cr
& {\text{Group like terms}} \cr
& 3{x^3} + 4{x^2} + 6x = \left( {A{x^3} + C{x^3}} \right) + \left( {A{x^2} + B{x^2} + 2C{x^2} + D{x^2}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { + 4Ax + Cx + 2Dx} \right) + 4A + 4B + D \cr
& {\text{We obtain the following system of equations}} \cr
& A + C = 3 \cr
& A + B + 2C + D = 4 \cr
& 4A + C + 2D = 6 \cr
& 4A + 4B + D = 0 \cr
& {\text{Solving using a calculator we obtain}} \cr
& A = 1,\,\,\,B = - 1,\,\,\,C = 2,\,\,\,D = 0 \cr
& {\text{Substituting the constants values into partial fr}}{\text{. decomposition}} \cr
& \frac{{3{x^3} + 4{x^2} + 6x}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 4} \right)}} = \frac{1}{{x + 1}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}} + \frac{{2x}}{{{x^2} + 4}} \cr
& {\text{Therefore,}} \cr
& \int {\frac{{3{x^3} + 4{x^2} + 6x}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 4} \right)}}} dx = \int {\frac{1}{{x + 1}}} dx - \int {\frac{1}{{{{\left( {x + 1} \right)}^2}}}dx + \int {\frac{{2x}}{{{x^2} + 4}}dx} } \cr
& = \ln \left| {x + 1} \right| + \frac{1}{{x + 1}} + \ln \left( {{x^2} + 4} \right) \cr
& = \frac{1}{{x + 1}} + \ln \left| {\left( {x + 1} \right)\left( {{x^2} + 4} \right)} \right| + C \cr} $$
