Cube rooted integral

**superphysics** · December 15th 2007, 03:31 AM

I need to solve the following question:

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

Any help/solution will be much appreciated.

**galactus** · December 15th 2007, 05:27 AM

Here's what Maple gave me:

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

**wingless** · December 15th 2007, 05:32 AM

I think this is not a solvable integral.

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

**galactus** · December 15th 2007, 05:48 AM

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.

**superphysics** · December 15th 2007, 06:01 AM

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

Some clever trick?

**topsquark** · December 15th 2007, 06:50 AM

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

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