Evaluate Integral with 3-x^4: Solution & Steps

[karush]

evaluate the integral
$
\displaystyle
\int
\frac{x}{\sqrt{3-x^4}} dx
$
it looks like a trig substation but with x in the denominator I did this

$
\displaystyle
\int
\left(3-x^4\right)^{-1/2}x\text{ dx}
$
with the $x/dx$ of ${3-x^4}=4x^3$ I couldn't see how to get how to get this to work
but the calc on this gives the right ans

 

[Jameson]

I think you can transform this into a standard integral form. If you use the substitution $u=x^2$ then $du=2xdx$ and the integral becomes:

$$\frac{1}{2}\int \frac{1}{\sqrt{3-u^2}}du$$

 

[karush]

well that sure looks better
will have to get back to this I ran out of time

MHB is really a great help.;)

 

[karush]

I think you can transform this into a standard integral form. If you use the substitution $u=x^2$ then $du=2xdx$ and the integral becomes:

$$\frac{1}{2}\int \frac{1}{\sqrt{3-u^2}}du$$

$\displaystyle
\frac{1}{2}
\int \frac{1}{\sqrt{3-u^2}}du
=
\frac{1}{2}
\left[\sin^{-1}\left(\frac{\sqrt{3}u}{3}\right)\right]
$
replacing $u$ with $x^2$
$\displaystyle
\frac{1}{2}
\sin^{-1}\left(\frac{\sqrt{3}x^2}{3}\right) + C
$