- #1

Badgerspin

- 15

- 0

Well, I'm stuck on one. I'm being asked for the values of A, B, and C for the following problem.

∫((9x^2+13x-83)/((x-3)(x^2 + 16)))dx

I can get it worked down pretty far, but I'm continually given back that I've done it incorrectly.

Broken down:

(9x^2+13x-83)/((x-3)(x^2 + 16)) = A/(x-3) + (bx + c)/(x^2 + 16)

From there, I get (multiplying both sides by the common denominator, and now only working with the right side as it's the only one that needs to be worked):

A(x^2+12) + Bx(x-3) + C(x-3)

= ax^2 + 12A + Bx^2 - 3Bx + Cx - 3C

= (A+B)x^2 + (C-3B)x + (12A - 3C)

Therefor:

-9x^2 + 13x -83 = (A+B)x^2 + (C-3B)x + (12A - 3C)

Breaking it down further, I get:

A + B = -9

3B - C = 13

12A - 3C = -83

I'm thinking something may be wrong with my values in the second part. Any help would be appreciated. Whatever answers I put in, I'm marked as incorrect. I went so far as to plug the last little part I posted into Wolfram|Alpha to confirm, and it spits the same answers I've been getting. I get the feeling that I did something incorrectly earlier on.