Good afternoon,

I have worked out the following problem and arrived at the following answer. Would someone mind verifying this for me?

$\displaystyle

{\int{\frac{\sqrt{4-x^2}}{x^2}}dx

$

let $\displaystyle x=2sin\theta$

let $\displaystyle dx=2cos\theta d\theta$

I arrive at

$\displaystyle

{\int{\frac{\sqrt{4-x^2}}{x^2}}

=\frac{1}{2}\ln\mid\frac{2+x}{\sqrt{4-x^2}}\mid+C$