Verified answer

∫x^2 sqrt(1+x^3) dx=

For the integrand x^2 sqrt(x^3+1), substitute u = x^3+1 and du = 3 x^2 dx:

= 1/3 ∫sqrt(u) du

The ∫of sqrt(u) is (2 u^(3/2))/3:

= (2 u^(3/2))/9+constant

Substitute back for u = x^3+1:

= 2/9 (x^3+1)^(3/2)+constant <---Final Answer

Hope this helps! :)

Will take a guess and ASSUME that you actually mean :-

I = ∫ x ² (x ³ + 1 )^(1/2) dx

Let u = x ³ + 1

du = 3 x ² dx

du/3 = x ² dx

I =(1/3) ∫ u^(1/2) du

I = (2/9) u^(3/2) + C

I = (2/9) ( x ³ + 1 )^(3/2) + C

Let z = (x³ + 1), dz = (3x² dx), (1/3)dz = x² dx

Integral becomes

∫ (1/3) √z dz

= (1/3) ∫ z^(1/2) dz

= (1/3) (2/3)z^(3/2) + C

= (2/9)(x³ + 1)^(3/2) + C