# Partial Fraction Integral

• Jul 19th 2008, 05:54 PM
tcRom
Partial Fraction Integral
Here's the problem, I keep getting stuck on one part...
$
\int\frac{x^2+2x}{(x^2+4)^2}~dx
$

After separating into 2 integrals and taking the partial fraction of the first part, I end up getting stuck on the following integral:

$
-4\int\frac{1}{(x^2+4)^2}~dx
$

My question at this point: what's my substitution? I'm not seeing it but know, based on form, that it has something to do with arctan. I picked up the answer from the Mathematica Integrator but need to know the steps.

$
-4\bigg(\frac{x}{8(x^2+4)}+\frac{1}{16}arctan\bigg[\frac{x}{2}\bigg]\bigg)
$

I'm in a 5 week Calc2 course and I feel like this guy --> (Headbang) after just the first week.
• Jul 19th 2008, 06:15 PM
Mathstud28
Quote:

Originally Posted by tcRom
Here's the problem, I keep getting stuck on one part...
$
\int\frac{x^2+2x}{(x^2+4)^2}~dx
$

After separating into 2 integrals and taking the partial fraction of the first part, I end up getting stuck on the following integral:

$
-4\int\frac{1}{(x^2+4)^2}~dx
$

My question at this point: what's my substitution? I'm not seeing it but know, based on form, that it has something to do with arctan. I picked up the answer from the Mathematica Integrator but need to know the steps.

$
-4\bigg(\frac{x}{8(x^2+4)}+\frac{1}{16}arctan\bigg[\frac{x}{2}\bigg]\bigg)
$

I'm in a 5 week Calc2 course and I feel like this guy --> (Headbang) after just the first week.

Let $x=2\tan(\theta)$

Try that
• Jul 19th 2008, 07:19 PM
Mathstud28
Quote:

Originally Posted by tcRom
Here's the problem, I keep getting stuck on one part...
$
\int\frac{x^2+2x}{(x^2+4)^2}~dx
$

After separating into 2 integrals and taking the partial fraction of the first part, I end up getting stuck on the following integral:

$
-4\int\frac{1}{(x^2+4)^2}~dx
$

My question at this point: what's my substitution? I'm not seeing it but know, based on form, that it has something to do with arctan. I picked up the answer from the Mathematica Integrator but need to know the steps.

$
-4\bigg(\frac{x}{8(x^2+4)}+\frac{1}{16}arctan\bigg[\frac{x}{2}\bigg]\bigg)
$

I'm in a 5 week Calc2 course and I feel like this guy --> (Headbang) after just the first week.

Sorry, I had to shave (Smirk).

Ok, so here is what we have

$\int\frac{dx}{(x^2+4)^2}$

And here is what we do

Let

$x=2\tan(\theta)$

$\Rightarrow{dx=2\sec^2(\theta)~d\theta}$

So we have

$\int\frac{dx}{(x^2+4)^2}$

$\underbrace{\Rightarrow}_{x=2\tan(\theta)}\int\fra c{2\sec^2(\theta)}{(4\tan^2(\theta)+4)^2}~d\theta$

$=\frac{1}{8}\int\frac{\sec^2(\theta)}{\sec^4(\thet a)}~d\theta$

$=\frac{1}{8}\int\cos^2(\theta)~d\theta$

$=\frac{1}{16}\int\bigg[1+\cos(2\theta)\bigg]~d\theta$

$=\frac{1}{16}\bigg[\theta+\frac{1}{2}\sin(2\theta)\bigg]+C$

$=\frac{1}{16}\bigg[\theta+\sin(\theta)\cos(\theta)\bigg]$

$\underbrace{=}_{\text{backsub}}\frac{1}{16}\bigg[\arctan\left(\frac{x}{2}\right)+\sin\left(\arctan\ left(\frac{x}{2}\right)\right)\cos\left(\arctan\le ft(\frac{x}{2}\right)\right)\bigg]+C$

$=\frac{1}{16}\bigg[\arctan\left(\frac{x}{2}\right)+\frac{x}{\sqrt{x^2 +4}}\cdot\frac{2}{\sqrt{x^2+4}}\bigg]+C$

$=\frac{1}{16}\bigg[\arctan\left(\frac{x}{2}\right)+\frac{2x}{x^2+4}\b igg]+C$
• Jul 19th 2008, 08:06 PM
tcRom
It's apparent that I've stepped into your kingdom.

Thank you very much for the help! (Rofl)