# Evaluate $\int \frac{\sqrt{1-4x^2}}{x} dx$

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

• 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 $v-1$. Dec 1, 2016 at 16:56

Let $\sqrt{1-4x^2}=u\implies1-4x^2=u^2\implies-4x\ dx=\ du$
$$\int\dfrac{\sqrt{1-4x^2}}x\ dx=\int\dfrac{\sqrt{1-4x^2}}{4x^2}(4x\ dx)$$
with the substitution from above we get $$dx=\frac{u}{-4x}du$$ and our integral will be $$-\int \frac{u^2}{1-u^2}du$$