∫ dx/ (x^2+9)^2 dx
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?
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))
=...
You are on the right track, but there are a few errors in your calculation. Let's go through the steps together to correctly evaluate the integral.
Given ∫ (1/(x^2+9)^2) dx, you correctly substituted x = 3tan(u) and found dx = 3sec^2(u) du. Substituting these values, we have:
I = ∫ (3sec^2(u) du) / (9tan^2(u)+9)^2
Now, let's simplify the denominator:
9tan^2(u) + 9 = 9(tan^2(u) + 1) = 9sec^2(u)
Substituting back into the equation:
I = ∫ (3sec^2(u) du) / ((9sec^2(u))^2)
= ∫ (3sec^2(u) du) / (81sec^4(u))
= (1/27) ∫ (sec^2(u) du) / (sec^2(u))^2
= (1/27) ∫ (sec^2(u) du) / sec^4(u)
= (1/27) ∫ sec^2(u) du / sec^4(u)
= (1/27) ∫ du / sec^2(u)
= (1/27) ∫ cos^2(u) du
Next, we can use the identity cos^2(u) = (1 + cos(2u))/2:
I = (1/27) ∫ (1 + cos(2u))/2 du
= (1/54) ∫ (1 + cos(2u)) du
= (1/54) (u + (1/2)sin(2u)) + C
Finally, substitute back u = arctan(x/3):
I = (1/54) (arctan(x/3) + (1/2)sin(2arctan(x/3))) + C
So, your ultimate result is almost correct. Just remember to substitute back u = arctan(x/3) instead of u = 3tan(u).