Answer
$$ - \frac{1}{{3a\left( {a{x^3} + b} \right)}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{{{\left( {a{x^3} + b} \right)}^2}}}} dx \cr
& {\text{substitute }}u = a{x^3} + b,{\text{ }}du = 3a{x^2}dx,{\text{ }}\frac{1}{{3a}}du = {x^2}dx \cr
& = \int {\frac{{{x^2}}}{{{{\left( {a{x^3} + b} \right)}^2}}}} dx = \int {\frac{{\left( {1/3a} \right)}}{{{u^2}}}} du \cr
& = \frac{1}{{3a}}\int {{u^{ - 2}}du} \cr
& {\text{find antiderivative}} \cr
& = \frac{1}{{3a}}\left( {\frac{{{u^{ - 1}}}}{{ - 1}}} \right) + C \cr
& = - \frac{1}{{3au}} + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{{3a\left( {a{x^3} + b} \right)}} + C \cr} $$
