How to evaluate ∫ 1/(x^2·sqrt(4 - x^2)) dx

Hi, I'm having trouble with the following question and would appreciate some help.

∫ 1/(x^2·sqrt(4 - x^2)) dx

My attempt is this:

let u = sqrt(4 - x^2), x^2 = 4 - u^2, u' = -x/u = sqrt(4 - u^2)/u;

=> ∫ 1/(u·(4 - u^2)) dx


Thanks.

Using your substitution,
x^2 = 4 - u^2
=> 2x dx = - 2u du
=> dx = - (u/x) du = - [u/√(4 - u^2)] du

=> integral
= - ∫ [u/√(4 - u^2)] du / u * (4 - u^2)
= - ∫ du/(4 - u^2) ^(3/2)

I will try as under.
Let x = 2sinu
=> dx = 2cosu du
=> Integral
= ∫ 2cosu du / [4sin^2 u * √(4 - 4sin^2 u)]
= (1/4) ∫ cosec^2 u du
= - (1/4) cotu + c
= - (1/4) cosu / sinu + c
= - (1/4) √(1 - sin^2 u) / sinu + c
= - (1/4) √(1 - x^2/4) / (x/2) + c
= - √(4 - x^2) / (4x) + c.