# Math Help - Partial Fraction Integral

1. ## Partial Fraction Integral

Hello, I need help integrating a rational function using partial fractions. I'm not sure how to make an integral sign on the computer or w/e, but its:

integral of (x-3)/((x^2+2x+4)^2) dx

I've tried using the method for repeated irreducible quadratic factors, but it seems to return the same function. Then I tried substituting u=x+1 and some other things but they didn't seem to get me anywhere; I tried splitting the numerator into 2 different integrals but I can't seem to figure it out that way either. Any help would be appreciated, thanks.

2. ## Re: Partial Fraction Integral

nasty one ... go to the Wolfram link, click "show steps"

integral (x-3)/(x^2+2x+4)^2 dx - Wolfram|Alpha

3. ## Re: Partial Fraction Integral

Originally Posted by kerouacan
Hello, I need help integrating a rational function using partial fractions. I'm not sure how to make an integral sign on the computer or w/e, but its:

integral of (x-3)/((x^2+2x+4)^2) dx

I've tried using the method for repeated irreducible quadratic factors, but it seems to return the same function. Then I tried substituting u=x+1 and some other things but they didn't seem to get me anywhere; I tried splitting the numerator into 2 different integrals but I can't seem to figure it out that way either. Any help would be appreciated, thanks.
\displaystyle \begin{align*} \frac{x - 3}{\left(x^2 + 2x + 4\right)^2} &= \frac{1}{2}\left[\frac{2x - 6}{\left(x^2 + 2x + 4\right)^2}\right] \\ &= \frac{1}{2}\left[\frac{2x + 2}{\left(x^2 + 2x + 4\right)^2} - \frac{8}{\left(x^2 + 2x + 4\right)^2}\right] \\ &= \frac{1}{2}\left[\frac{2x + 2}{\left(x^2 + 2x + 4\right)^2}\right] - 4\left\{ \frac{1}{\left[(x + 1)^2 + 3\right]^2} \right\} \end{align*}

The first of those terms can be integrated using the substitution \displaystyle \begin{align*} u = x^2 + 2x + 4 \end{align*}. The second you'll need to substitute \displaystyle \begin{align*} x + 1 = \sqrt{3}\tan{\theta} \end{align*}.

4. ## Re: Partial Fraction Integral

Originally Posted by Prove It
\displaystyle \begin{align*} \frac{x - 3}{\left(x^2 + 2x + 4\right)^2} &= \frac{1}{2}\left[\frac{2x - 6}{\left(x^2 + 2x + 4\right)^2}\right] \\ &= \frac{1}{2}\left[\frac{2x + 2}{\left(x^2 + 2x + 4\right)^2} - \frac{8}{\left(x^2 + 2x + 4\right)^2}\right] \\ &= \frac{1}{2}\left[\frac{2x + 2}{\left(x^2 + 2x + 4\right)^2}\right] - 4\left\{ \frac{1}{\left[(x + 1)^2 + 3\right]^2} \right\} \end{align*}

The first of those terms can be integrated using the substitution \displaystyle \begin{align*} u = x^2 + 2x + 4 \end{align*}. The second you'll need to substitute \displaystyle \begin{align*} x + 1 = \sqrt{3}\tan{\theta} \end{align*}.
Thanks a lot guys, all those steps were in my book, I guess I just couldn't put them together.