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I'm struggling to figure out where i am going wrong with this integral. Any help would be much appreciated. Attached below my attempt at the problem:

Problem attempt

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  • $\begingroup$ You made a mistake with the partial fraction decomposition near the end. Both your denominators as $v+1$ and that ain't right. One should be $v+1$ and the other $v-1$. $\endgroup$ – imranfat Dec 1 '16 at 16:56
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HINT:

Let $\sqrt{1-4x^2}=u\implies1-4x^2=u^2\implies-4x\ dx=\ du$

$$\int\dfrac{\sqrt{1-4x^2}}x\ dx=\int\dfrac{\sqrt{1-4x^2}}{4x^2}(4x\ dx)$$

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with the substitution from above we get $$dx=\frac{u}{-4x}du$$ and our integral will be $$-\int \frac{u^2}{1-u^2}du$$

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