# Regarding partial fractions

hi, the problem is:

&int; (8x-17)/x^2+x-12 dx = &int; .../(x+4)(x-3) dx

so 8x - 17 = A(x-3) + B(x+4)

(A+B)x + 4B - 3A

so we have 2 eq and 2 unknown

A+B = 8
4B-3A = -17

... but the book says it's suppose to be 4A - 3B,.. I don't know what I did wrong.

jamesrc
Gold Member
They probably just defined A and B differently than you did. (They must have said (8x-17)/(x2+x-12) = A/(x-3) + B/(x+4) ). You'll still get the right answer your way.

the ending result is different...

they got B = 7 and A = 1...

I got B = 41/7, and A = 15/7

Hurkyl
Staff Emeritus
Gold Member
Well, the B and A you listed aren't solutions to the equations you got...

Tom Mattson
Staff Emeritus
Gold Member
Originally posted by PrudensOptimus
(A+B)x + 4B - 3A

so we have 2 eq and 2 unknown

A+B = 8
4B-3A = -17

You don't have to solve these like that.

so 8x - 17 = A(x-3) + B(x+4)

Starting from here, just plug in x=+3 and x=-4.

x=3:
8(3)-17=A(3-3)=B(3+4)
7=7B
B=1

x=-4:
8(-4)-17=A(-4-3)+B(-4+4)
-49=-7A
A=7

amazing.