`int x^3sqrt(x^2 + 1) dx` Evaluate the indefinite integral.

You need to use the following substitution ` x^2 + 1=u` , such that:

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

`x^2 = u-1`

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

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

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

`(1/2) int (u-1)*u^(1/2)*du = (1/5) u^(5/2) - (1/3) u^(3/2) + c`

Replacing back `x^2 + 1` for u yields:

`int x^3*sqrt(x^2 + 1)dx = (1/5) (x^2 + 1)^(5/2) - (1/3) (x^2 + 1)^(3/2) + c `

Hence, evaluating the indefinite integral, yields `int x^3*sqrt(x^2 + 1)dx = (1/5) (x^2 + 1)^(5/2) - (1/3) (x^2 + 1)^(3/2) + c`