View Full Version : Regarding partial fractions
PrudensOptimus
Nov15-03, 02:49 PM
hi, the problem is:
∫ (8x-17)/x^2+x-12 dx = ∫ .../(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.
Please help.
jamesrc
Nov15-03, 02:55 PM
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.
PrudensOptimus
Nov15-03, 03:19 PM
the ending result is different...
they got B = 7 and A = 1...
I got B = 41/7, and A = 15/7
Well, the B and A you listed aren't solutions to the equations you got...
Tom Mattson
Nov15-03, 05:42 PM
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
PrudensOptimus
Nov15-03, 07:48 PM
amazing.
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