# Calculus- Integration w/ partial fractions

• February 10th 2009, 07:16 PM
juicysharpie
Calculus- Integration w/ partial fractions
Find the integral of (x^2+4)/(x^2-4) dx.

The answer is x + 2 ln | x-2/x+2 |

Here's what I did so far:
(x^2+4)/(x^2-4) = (x^2+4)/[x+2][x-2].
I'm not really sure how to split them apart. I tried making it Ax+B/x+2 + Cx+D/x-2 but when I worked it out I got: A+C = 1, C-B = 2..? I don't think that was right. (Shake)

• February 10th 2009, 07:23 PM
skeeter
$\frac{x^2 + 4}{x^2 - 4} = \frac{x^2 - 4 + 8}{x^2 - 4} = 1 + \frac{8}{x^2-4}$

partial fraction set up for last quotient ...

$\frac{8}{x^2-4} = \frac{A}{x+2} + \frac{B}{x-2}$

finish
• February 10th 2009, 08:02 PM
Jhevon
Just for fun: $\frac 8{(x + 2)(x - 2)} = \frac {2x + 4 - (2x - 4)}{(x + 2)(x - 2)} =$ $\frac {2x + 4}{(x + 2)(x - 2)} - \frac {2x - 4}{(x + 2)(x - 2)} = \frac 2{x - 2} - \frac 2{x + 2}$
• February 11th 2009, 08:59 AM
Krizalid
Quote:

Originally Posted by Jhevon

Just for fun: $\frac 8{(x + 2)(x - 2)} = \frac {2x + 4 - (2x - 4)}{(x + 2)(x - 2)} =$ $\frac {2x + 4}{(x + 2)(x - 2)} - \frac {2x - 4}{(x + 2)(x - 2)} = \frac 2{x - 2} - \frac 2{x + 2}$

Not fun!! Actually better and faster! (Sun)
• February 11th 2009, 11:30 AM
Mathislife
Quote:

Originally Posted by Jhevon
Just for fun: $\frac 8{(x + 2)(x - 2)} = \frac {2x + 4 - (2x - 4)}{(x + 2)(x - 2)} =$ $\frac {2x + 4}{(x + 2)(x - 2)} - \frac {2x - 4}{(x + 2)(x - 2)} = \frac 2{x - 2} - \frac 2{x + 2}$

wow never would've thought of it