How can I solve this tricky integral involving a radical and trig substitution?

[krusty the clown]

\int \frac {x}{\sqrt{3-x^4}}\mbox{dx}

Basically I haven't had any luck with anything I have tried. It almost looks like a trig substitution, but I don't know how to deal with the x^4 under the radical. Also, I would rather have a hint as to where to start rather than have the whole problem done for me.


-Thanks, Erik

 

[Pyrrhus]

Deal it as (x^2)^2

 

[Tide]

I think that's going to turn into an incomplete beta function.

 

[krusty the clown]

I think that's going to turn into an incomplete beta function.

I'm not sure what this means, but this problem is from calc II, and the text specifically says there are answers to all of the questions from the section this was taken from.

Deal it as (x^2)^2

Thanks. I am prety sure I can do this with trig substituion now, but if I can't I will let you know.


-Thanks, Erik

 

[Tide]

Right! I didn't look at it long enough. It's going to be an arcsin with an x^2 in the argument.

 

[Pyrrhus]

Read Tide's post he gave it away
No need for trigonometric substitution.

Hint!
u = x^2
\frac{du}{2}= xdx