Integration of 1/(x^2+4)^2 with trigonometric substitution

[badtwistoffate]

Int[1/(x^2+4)^2]
let x= 2 tan(q)
i get 1/(4 tan^2(q) + 4)^2
I get 1/ (16 tan ^4(q) + 32 (1-sec^2(q) + 16)
and now i think about doing a u-substituition with tan x=u, du =sec^2 (x) can i do that!?!

 

[badtwistoffate]

do you just don't like my topics i want help in?

 

[whozum]

$$\frac{1}{(4\tan(q)+4)^2} = \frac{1}{16\sec^4q}$$