Question

How do you integrate `x^5 * sqrt(x^3+1) dx`?

Answer 1

The integral is `2/15(x^3 + 1)^(5/2) - 2/9(x^3 + 1)^(3/2) + C`

Explanation:

Let `u = x^3 + 1`. Then `du = 3x^2dx -> dx =(du)/(3x^2)`. Let `I` be the integral.

`I= intx^5sqrt(u) * (du)/(3x^2)`

`I = 1/3intx^3sqrt(u)du`

`I = 1/3int(u - 1)sqrt(u)du`

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

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

`I = 1/3intu^(3/2)du - 1/3intu^(1/2)du`

`I = 2/5(1/3)u^(5/2) - 2/3(1/3)u^(3/2) + C`

`I = 2/15u^(5/2) - 2/9u^(3/2) +C`

`I = 2/15(x^3 + 1)^(5/2) - 2/9(x^3 + 1)^(3/2) + C`

Hopefully this helps!