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

1 Answer
May 2, 2017

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!