`int x^2 sqrt(2 + x) dx` Evaluate the indefinite integral.

You need to evaluate the indefinite integral `int x^2*sqrt(2+x) dx` using the following substitution  `2+x=u` , such that:

`2+x=u=>dx = du `

`x = u-2`

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

`int (u-2)*sqrt(u) dx= (2/5)u^(5/2) - (4/3)u^(3/2) + c`

Replacing back `2+x ` for u yields:

`int x^2*sqrt(2+x) dx = (2/5)(2+x)^(5/2) - (4/3)(2+x)^(3/2) + c`

Hence, evaluating the indefinite integral, yields  `int x^2*sqrt(2+x) dx = (2/5)(2+x)^(5/2) - (4/3)(2+x)^(3/2) + c`