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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!

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Good try with the $\LaTeX$ formatting. Just some tips for next time: When you use code such as \int or \frac, don't put a space after them. Also, don't forget to surround your math with $ $; this formats the math.You can see how I coded your post if you right click the formatted code and click "Show Math As"> "Tex Commands". -Regards – Sujaan Kunalan May 1 '14 at 15:59
@SujaanKunalan It is a good idea to use a space after \int, as \intx\;dx will give an error, whereas \int x\;dx will not. (But I agree regarding \frac) – apnorton May 1 '14 at 16:01
Yes of course, I forgot about \int x . Thanks. – Sujaan Kunalan May 1 '14 at 16:08
@SujaanKunalan Why does it matter? The spaces don't do anything. – Akiva Weinberger Nov 3 '14 at 22:41
@columbus8myhw: It matters because $\intx$ results in an error, but $\int x$ formats the integral sign properly. – Sujaan Kunalan Nov 4 '14 at 2:58

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.

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$$\int\frac{x\,dx}{\sqrt{4-3x^4}} = \int\frac{x\, dx}{\sqrt3\sqrt{\frac43-x^4}}$$

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?

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$$\int\frac{x\ dx}{\sqrt{4-3x^4}}=\int\frac{x^3\ dx}{x^2\sqrt3\sqrt{\frac43-x^4}}$$

Set $x^2=u$ or $\sqrt{\frac43}\sin\theta$

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  1. get rid of the fraction ...

$$1/x = x^-1$$

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

  1. get rid of the square root ...

$$\sqrt{x} = x^{1/2}$$

... so ...

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

  1. 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.)

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