Question

The integral of x/(sqrt(3-x^4)dx

Answer 1

\[\int\limits_{}^{} x/\sqrt{3-x^4}dx\]

Answer 2

Well, let u=x^2
Then du=2xdx
(1/2)du=xdx
Your integral becomes:
\[\frac{1}{2}\int\limits \frac{u}{\sqrt{3-u^2}}du\]
From here you can use trig sub. Do you need me to take it further?

Answer 3

I thought I wouldn't be able to use trig sub because of the 3...

Answer 4

Of course you can. It doesn't have to be a 1. Or a perfect square for that matter.
You have:
\[\sqrt{(\sqrt{3})^2-u^2}\]
Use sin(theta) for your trig sub.

Answer 5

Oh, okay I thought you had to have a number that was a perfect square :)

Answer 6

Oh and quick question, how does the x on top become u?

Answer 7

It shouldn't be. I just mistyped it.

Answer 8

It should be just du.

Answer 9

Tell me if you get stuck :P

Answer 10

So the x on top gets cancelled out?

Answer 11

Yeah, when you replace it with du. :)

Answer 12

Oh yea, okay I see once I wrote it down :) Thank you!