# Integrating 1/(4+x^2)^2 dx

1. Oct 11, 2009

### oexnorth

I'm supposed to be Integrating 1/(4+x^2)^2 dx using trigonometric substitutions, but I do not know how to get started with this one.

Am I supposed to rewrite this as 1/sqrt((4+x^2)^2) and then use the reference triangle? I've tried expanding the bottom, but that does'nt get me anywhere.

Any help, please. I don't want the answer, I just want the first step to get this started.
Thanks

2. Oct 11, 2009

### tiny-tim

Welcome to PF!

Hi oexnorth! Welcome to PF!
With (a2 - x2), use sine or cosine.

With (x2 - a2), use sec or cosec.

With (a2 + x2), use tan or cot.

3. Oct 11, 2009

### oexnorth

So would x=2sqrt(tan(theta))? Sorry, but the square instead of the square root in the problem is throwing me off.

4. Oct 11, 2009

### Mentallic

Yes exactly.
You're using the fact that $$1+tan^2\theta=sec^2\theta$$
so I'm sure you can see where this is heading.

p.s. do not start throwing in square roots. That is an invalid move, and makes the problem boring now with the invention of the standard integrals and all

5. Oct 11, 2009

### oexnorth

Definitely. Thank you both.

6. Oct 11, 2009

### tiny-tim

(have a theta: θ )
eugh :yuck: … you can use √tan, but it's easy to make a mistake if you choose anything complicated …

I'd strongly recommend sticking to the simplest possible substitution, in this case just tan.

7. Oct 11, 2009

### oexnorth

Ok, after repeated workings and reworkings, I've come up with:

(1/16)arctan(x/2)+x/8(4+x^2)

Can anyone verify this? Thanks.

8. Oct 11, 2009

### Staff: Mentor

You can verify that it's correct by differenting. If you answer is correct, it should be that d/dx[1/16*arctan(x/2) + 1/8*x(x^2 + 4)] = 1/(x^2 + 4)^2.

Whatever your answer turns out to be, don't forget the constant of integration.