Question

use partial fractions to find.

Answer 1

\[\int\limits \frac{1}{x^2-9}\]

Answer 2

\[\int\limits \frac{1}{(x-3)(x+3)}\]

Answer 3

hmm if there was something besides 1 on top this would be eaiser....

Answer 4

Idk how to get A+ B from this.... unlessI do 1 + 0...

Answer 5

\[\frac{A}{x-3} + \frac{B}{x+3} = \frac{1}{(x+3)(x-3)}\]
\[A(x+3) + B(x-3) = 1\]
you know what i did right?

Answer 6

oh I guessI didn't need to doanyhing special for this then... Yes I do sir :D

Answer 7

if you set x = 3 then A(6) = 1 A = 1/6

Answer 8

if you set x =-3 then B(-6) =1 B = -1/6

Answer 9

so now it becomes 1/6/(x-3) + -(1/6)/(x-3)

Answer 10

yup \[\frac{1}{6(x-3)} - \frac{1}{6(x+3)}\]
now you integrate that

Answer 11

\[\frac 16 \int \frac{1}{x-3} dx - \frac 16 \int \frac{1}{x+3} dx\]

Answer 12

1/6 ln|x-3| - 1/6 ln|x+3| +c :)

Answer 13

yup