Asked by chrystabelle
express x-1/(x+1)(x-2)^2 into partial fractions
Answers
Answered by
Reiny
let
(x-1)/((x+1)(x-2)^2) = A/(x-1) + B/(x-2) + C/(x-2)^2
multiply both sides by (x+1)(x-2)^2
A(x-2)^2 + B(x-2)(x+1) + C(x+1) = x-1
let x = 2, ---> 3C = 1 or C=13
let x = -1 ---> 9A + -2 or A = -2/9
let x = 0 ----> 4A - 2B + C = -1
4(-2/9) - 2B + 1/3 = -1
times 9
-8 - 18B + 3 = -9
B = 2/9
so (x-1)/(x-2)^2 = -2/(9(x+1)) + 2/(9(x-2)) + 1/(3(x-2)^2)
verified with Wolfram:
http://www.wolframalpha.com/input/?i=%28x-1%29%2F%28%28x%2B1%29%28x-2%29%5E2%29
(x-1)/((x+1)(x-2)^2) = A/(x-1) + B/(x-2) + C/(x-2)^2
multiply both sides by (x+1)(x-2)^2
A(x-2)^2 + B(x-2)(x+1) + C(x+1) = x-1
let x = 2, ---> 3C = 1 or C=13
let x = -1 ---> 9A + -2 or A = -2/9
let x = 0 ----> 4A - 2B + C = -1
4(-2/9) - 2B + 1/3 = -1
times 9
-8 - 18B + 3 = -9
B = 2/9
so (x-1)/(x-2)^2 = -2/(9(x+1)) + 2/(9(x-2)) + 1/(3(x-2)^2)
verified with Wolfram:
http://www.wolframalpha.com/input/?i=%28x-1%29%2F%28%28x%2B1%29%28x-2%29%5E2%29
Answered by
Reiny
in the middle of the solution above the line should read:
et x = 2, ---> 3C = 1 or C=1/3
et x = 2, ---> 3C = 1 or C=1/3
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