#### Answer

$$\frac{1}{6}\ln 3$$

#### Work Step by Step

$$\eqalign{
& \int_0^1 {\frac{{{x^2}dx}}{{2{x^3} + 1}}} \cr
& {\text{use substitution}}{\text{. Let }}u = 2{x^3} + 1,{\text{ so that }}\frac{{du}}{{dx}} = 6{x^2},\,\,\,\,\,{x^2}dx = \frac{1}{6}du \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = 2{\left( 1 \right)^3} + 1 = 3 \cr
& \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = 2{\left( 0 \right)^3} + 1 = 1 \cr
& {\text{Then}} \cr
& \int_0^1 {\frac{{{x^2}dx}}{{2{x^3} + 1}}} = \int_1^3 {\frac{{\left( {1/6} \right)du}}{u}} \cr
& = \frac{1}{6}\int_1^3 {\frac{{du}}{u}} \cr
& {\text{integrate}} \cr
& = \frac{1}{6}\left( {\ln \left| u \right|} \right)_1^3 \cr
& {\text{evaluating}} \cr
& = \frac{1}{6}\left( {\ln \left| 3 \right| - \ln \left| 1 \right|} \right) \cr
& = \frac{1}{6}\ln 3 \cr} $$