Answer
$\dfrac{52}{9}$
Work Step by Step
Given: $I=\int_0^2 x^2 \sqrt {x^3+1} \ dx$
Let us consider that $u=x^3+1 \implies dx=\dfrac{du}{3x^2}$
Now, we have $I=\int_0^2 x^2 u^{1/2} (\dfrac{du}{3x^2})$
or, $=\dfrac{1}{3} \int_{0}^2 u^{1/2} \ du$
or, $=\dfrac{1}{3} [\dfrac{u^{3/2}}{3/2}]_{0}^{2} +C$
or, $=\dfrac{2}{9} [(x^3+1)^{3/2}]_{0}^2 $
or, $=\dfrac{2}{9} [(2^3+1)^{3/2}]-\dfrac{2}{9} [(0+1)^{3/2}]$
or, $=\dfrac{52}{9}$
