Question

Calculus 2,
Help me solve this
http://screencast.com/t/LaeHMSSqKak

Answer 1

I believe you need integration by parts. Have you learned how to do that?

Answer 2

i guess splitting this into partial fractions might help

Answer 3

Oh yep that looks like a much better idea.

Answer 4

So I think the first step would be to factor the denominator into x(x+3)(x-3)

Answer 5

then ?

Answer 6

yes

Answer 7

\[\frac{A}{x} + \frac{B}{x+3}+\frac{C}{x-3} = \frac{2x^{2}-9x-9}{x(x+3)(x-3)}\]

Answer 8

thx a lot guys

Answer 9

So \[A+B+C=2x^2-9x-9\]

Answer 10

can I do that ??

Answer 11

no

Answer 12

ok

Answer 13

Oh no, I'm sorry :/ Been way too long...I swear that's how it was done though. Guess I'm too rusty!

Answer 14

hehe np man, its the try that matters ? :p

Answer 15

! *

Answer 16

Oh, I think \[A(x^2-9)+B(x^2-3x)+C(x^2+3x)=2x^2-9x-9\]

Answer 17

That makes much more sense.

Answer 18

Then you can say \[Ax^2+Bx^2+Cx^2=2x^2\]

Answer 19

And do the same for the x term and the constant term. Then using those three equations, solve for A, B, and C.

Answer 20

ohh ty!!

Answer 21

yes @escolas :)