Simplifying a u-substitution for $\int \frac{x} { \sqrt {4-3 x^4 } } \, dx$ this is a calculus one problem I cannot figure out.  I may be making a simple assumption in my substitutions, please help.  (I hope I typed this correctly, this is my first time using the MathJaX formatting.)  Thanks!
$\int \frac{x} { \sqrt {4-3 x^4 } } \, dx$
I let $u = 3 x^4$, then $du = 12 x^3$.  I then used $\sqrt{u} = \sqrt{3} x^2$.  When I substituted in I got the the following integral which I can't figure out how to simplify:
$\frac{1}{12}  {\int \frac{\sqrt{3}} {\sqrt{u} \sqrt{4-u}}} \, du$
Thanks for any help at all!
 A: Let us at first not try to make the "best" substitution, just something that makes things look better.  Let $u=x^2$. Why this? Because the derivative of $x^2$ is basically sitting on top, and the rest is a function of $x^2$. 
Thus $x\,dx=\frac{1}{2}\,du$. Our integral becomes $\int \frac{1}{2}\cdot\frac{1}{\sqrt{4-3u^2}}\,du$. Not finished yet, but progress, you have probably seen something like this before, and know how to handle it.
Remark: In hindsight, we might make the substitution $\sqrt{3}\,x^2=2u$, and then the thing collapses immediately. 
A: $$\int\frac{x\,dx}{\sqrt{4-3x^4}} = \int\frac{x\, dx}{\sqrt3\sqrt{\frac43-x^4}}$$
Try using trigonometric substitution: let $$x^2 = \sqrt {\frac 43} \sin \theta \implies 2x\,dx = \sqrt{\frac 43}\cos \theta$$
That gives us the integral $$\frac 23 \int \dfrac{\cos \theta \,d\theta}{\sqrt{1 - \cos^2 \theta}}$$
Can you take it from here?
A: $$\int\frac{x\ dx}{\sqrt{4-3x^4}}=\int\frac{x^3\ dx}{x^2\sqrt3\sqrt{\frac43-x^4}}$$
Set $x^2=u$ or $\sqrt{\frac43}\sin\theta$
A: *

*get rid of the fraction ... 


$$1/x = x^-1$$
...so...
$$\frac {x}{\sqrt{4-3x^4}} = x\sqrt{4-3x^4}^{-1}$$


*get rid of the square root ...


$$\sqrt{x} = x^{1/2}$$
... so ...
$$x\sqrt{4-3x^4}^{-1} = x(4-3x^4)^{-1/2}$$


*Simplified version of u-substitution is ...


$$\frac {k}{u'} * \frac {u^{n+1}}{n+1}+C$$
... your variables are ...
$$k = x$$
$$u = 4-3x^4$$
$$u' = -12x^3$$
$$n = -1/2$$
$$n + 1 = -1/2 + 2/2 = 1/2$$
so...
$$\frac {x}{-12x^3} * \frac {u^{\frac 12}}{\frac{1}{2}} + C$$
...x on top cancels out, and dividing by 1/2 is same as multiplying by 2/1...
$$\frac {1}{-12x^2} * (2 * u^{\frac 12}) + C$$
... multiply it all together ...
$$\frac {2u^{\frac 12}}{-12x^2}+C$$
... simplify ...
$$\frac {u^{\frac 12}}{-6x^2}+C$$
... replace your u ...
$$\frac {(4-3x^4)^{\frac 12}}{-6x^2}+C$$
... flip your numerator back to a square root ...
$$\frac {\sqrt{4-3x^4}}{-6x^2}+C$$
... we can try simplifying some on the square root, since 4 is just $2^2$, and $\sqrt{x^4} = x^2$
$$\frac {\sqrt{2^2-3x^4}}{-6x^2}+C$$
$$\frac {2+x^2\sqrt{-3}}{-6x^2}+C$$
...since we have the $x^2$ out now, we can cancel with with the denominator...
$$\frac {2+\sqrt{-3}}{-6}+C$$
... square root of a negative number is an imaginary number, so you'd have to flip things around to get rid of it for further simplification. We've just simplified the x right out of the equation, though, so all that would be left is to solve for C (the constant).
(Side note, I'm in college and taking calculus now. We just got into u-substitution, and I figured out that in cases like the one above the formula distills down to k / u' * (u^n+1 / n+1).  "k" is any part outside the "u^n" ... eg: 600x(4x+1)^2 ... "k" = 600x, u = 4x+1, u' = 4).  Basically, in regards to simply u-substitution, you're taking parts of the function outside the u^n, and dividing them by the u'.  If there's no "k" part then k=1 ... eg: (3x^2 + 5)^7 ... is the same as 1 * (3x^2 + 5)^7 ... so k = 1 / u' ... or 1 / 6x in this case.)
