How do you integrate #\int \frac { x ^ { 2} + 2} { \root[ 3] { x ^ { 3} + 6x + 1} } d x#?

1 Answer
Jun 23, 2017

The integral equals #1/2(x^3 + 6x + 1)^(2/3) + C#

Explanation:

We let #u = x^3 + 6x + 1#, so #du = 3x^2 + 6 dx# and #dx = (du)/(3x^2 + 6)#.

#I = int (x^2 + 2)/root(3)(u)* (du)/(3(x^2 + 2))#

#I = 1/3 int 1/root(3)(u) du#

#I = 1/3int u^(-1/3)#

Now use #intx^n dx = (x^(n + 1))/(n + 1) + C#

#I = 1/2u^(2/3) + C#

#I = 1/2(x^3 + 6x + 1)^(2/3) + C#