`int x^2 sqrt(x^3 + 1) dx, u = x^3 + 1` Evaluate the integral by making the given substitution.

You need to evaluate the indefinite integral by performing the indicated substitution` u =x^3 + 1` , such that:

`u = x^3 + 1 => du = 3x^2 dx=> x^2dx = (du)/3`

`int x^2sqrt(x^3+1)dx = (1/3)*int sqrt u du`

Using the formula `int u^n du = (u^(n+1))/(n+1) + c` yields

`(1/3)*int sqrt u du = (1/3)(u^(1/2+1))/(1/2+1) + c`

`(1/3)*int sqrt u du = (2/9)(u^(3/2)) + c`

Replacing back  `x^3 + 1 ` for u yields:

`int x^2sqrt(x^3+1)dx = (2/9)((x^3 + 1)^(3/2)) + c`

Hence, evaluating the indefinite integral yields `int x^2sqrt(x^3+1)dx = (2/9)(x^3 + 1)sqrt(x^3+1) + c`