Asked by Mike
Solve the integral (x+3)/(x-1)^3
So far, I'm stuck. This is what I have thus far:
x+3 = A(x-1)^3/(x-1) + B(x-1)^3/(x-1)^2 + C(x-1)^3/(x-1)^3
x+3 = A(x^2-2x+1) + B(x-1) + C
x+3 = Ax^2 - 2Ax +Bx + A - B + C
x+3 =Ax^2 + 1(B-2A)x + (A - B + C)
So far, I'm stuck. This is what I have thus far:
x+3 = A(x-1)^3/(x-1) + B(x-1)^3/(x-1)^2 + C(x-1)^3/(x-1)^3
x+3 = A(x^2-2x+1) + B(x-1) + C
x+3 = Ax^2 - 2Ax +Bx + A - B + C
x+3 =Ax^2 + 1(B-2A)x + (A - B + C)
Answers
Answered by
scott
let u = x-1
(x+3)/(x-1)^3 = (u + 4) / u^3 = u^(-2) + 4 u^(-3)
(x+3)/(x-1)^3 = (u + 4) / u^3 = u^(-2) + 4 u^(-3)
Answered by
Mike
Okay, I understand how to solve now, but I'm curious on where did that four come from?
Thank you.
Thank you.
Answered by
Steve
distributive property: (u+4)*v = uv + 4v
duh
duh
Answered by
scott
x - 1 = u ... x + 3 = u + 4