How do I integrate a rational function with a square root in the denominator?

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\int \frac{x}{\sqrt(1-4x^{2})}dx
If it was just a constant on top I could turn it into an arcsin, but there is a stupid x there, so not to sure. I though about intergrating by parts, but their must be an easier method than that?

Thanks
 
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Well, you could make the x above "disappear" by doing a change of variable u=4x².
 
Try the substitution u=1-4x^2.
 
cheers thanks got it!

btw, how do you spot a substitution? There is a rule(s) for it. When somthingis the product or differential of somthing else?

Thanks :)
 
When you see a multiple of the derivative of something else in the numerator, it's a dead giveaway for a substitution.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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