Question

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

Answer 1

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`