∫ dx/ (x^2+9)^2 dx

Question

set x = 3tan u
dx = 3 sec^2 u du

I = 3 sec^2 u du / ( 9 tan^2 u + 9)^2
= 3 sec^2 u du / ( 81 ( tan^2 u + 1)^2
= sec^2 u du / ( 27 ( sec^2 u )^2
= du / ( 27 sec^2 u
= 2 cos^2 u du / 54
= ( 1 + cos 2u) du / 54
= ( u + sin 2u / 2) / 54
= ( arctan x/3 + sin u cos u ) / 54

Am I going in the right direction or did I totally screw up?

Answer 1

You're ok so far.
You only need to convert sin(u)cos(u)/54 back in terms of x, which should be x/(18x²+162).
You can start by
sin(u)cos(u)/54=cos^2(u)tan(u)/54
=3tan(u)/[162(sec²(u))
=...