# Thread: Cube rooted integral

1. ## Cube rooted integral

I need to solve the following question:

$\int{\sqrt[3]{x^2+1}} dx$

Any help/solution will be much appreciated.

2. Here's what Maple gave me:

$x\cdot\text{hypergeom}\left( [1/3,1/2],[3/2],-{x}^{2} \right)$

3. I think this is not a solvable integral.

Mathematica gave:
$\frac{1}{5} x \left(3 \left(1+x^2\right)^{1/3}+2 \text{Hypergeometric2F1}\left[\frac{1}{2},\frac{2}{3},\frac{3}{2},-x^2\right]\right)$

4. It does not have a nice indefinte form. If it had limits of integration, perhaps Kriz or someone could have a clever way to do it?. Here's the graph. At first glance, one would think it'd be easy.

5. Ok, if it's unsolvable indefinitely, can we do it in the range $0$ to $a$?

Some clever trick?

6. Originally Posted by superphysics
Ok, if it's unsolvable indefinitely, can we do it in the range $0$ to $a$?

Some clever trick?
What galactus meant was something like from 0 to $\infty$. You can't put this into a form with a variable.

You can put it in the form of an infinite series. (That's how the hypergeometric function is typically written.)

-Dan