How do you integrate #int 2x^2sqrt(x^3+1)dx# from [1,2]?

1 Answer
Jan 25, 2017

#12 - 8/9sqrt(2)#

Explanation:

This is a substitution problem. Let #u =x^3 + 1#. Then #du = 3x^2dx# and #dx = (du)/(3x^2)#.

#int_1^2 2x^2sqrt(u) * (du)/(3x^2)#

#int_1^2 2/3sqrt(u) du#

#2/3int_1^2 sqrt(u)du#

#2/3[2/3u^(3/2)]_1^2#

#2/3[2/3(x^3 + 1)^(3/2)]_1^2#

#2/3[2/3(2^3 + 1)^(3/2) - 2/3(1^3 +1)^(3/2)]#

#2/3[2/3(27) - 2/3sqrt(8)]#

#2/3[18 - 4/3sqrt(2)]#

#12 - 8/9sqrt(2)#

Hopefully this helps!