Answer
$$\frac{1}{6}\ln \left| {2{x^3} + 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}dx}}{{2{x^3} + 1}}} \cr
& \cr
& {\text{integrate by substitution}} \cr
& {\text{set }}u = 2{x^3} + 1{\text{ then }}\frac{{du}}{{dx}} = 6{x^2} \to {x^2}dx = \frac{1}{6}du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{{x^2}dx}}{{2{x^3} + 1}}} = \int {\frac{{\left( {1/6} \right)du}}{u}} \cr
& = \frac{1}{6}\int {\frac{{du}}{u}} \cr
& {\text{integrating}} \cr
& = \frac{1}{6}\ln \left| u \right| + C \cr
& {\text{replace }}2{x^3} + 1{\text{ for }}u \cr
& = \frac{1}{6}\ln \left| {2{x^3} + 1} \right| + C \cr} $$
