\({1_0^∫} x^2(x^3+2)^3\)dx

let \( u = x^3 + 2, du = 3x^2dx\)

when x = 1,  u = 3

when x = 0,  u = 2

dx = \(\frac{du}{3x^2}\)

\({3_2^∫}\) \(\frac{x^2[u]^3}{3x^2}\)

\({3_2^∫}\) \(\frac{u^3}{3}\) du 

= \(\frac{u^4}{3*4}\)\(_2\)3

\(\frac{1}{12} [u^4]\)\(_2\)3

\(\frac{1}{12} [3^4 - 2^4]\)

\(\frac{1}{12}[81 - 16]\)

\(\frac{65}{12}\)