Wolfram split it up into these 3 fractions:
http://www.wolframalpha.com/input/?i=partial+fractions+(2x-3)%2F(x%5E3-x%5E2)
notice, possible denominators could have been,
x, x^2, and (x-1)
How did you do your original split-up?
Solve the integral: (2x-3)/(x^3-x^2)
I'm stuck at finding A. I did partial fractions. I've already found B and C.
At x=1, C=-1
At x=0, B=3
2 answers
I did that.
∫A/x +∫B/x^2 +∫C/(x-1)
F(X)=A ln|x| + B/x + C ln|x-1| + C
(2x-3)/[x^2(x-1)] = [A(x^2)(x-1)]/[x] + [B(x^2)(x-1)]/[x^2] + [C(x^2)(x-1)]/[x-1]
2x-3 = A(x)(x-1) + B(x-1) + C(x^2)
∫A/x +∫B/x^2 +∫C/(x-1)
F(X)=A ln|x| + B/x + C ln|x-1| + C
(2x-3)/[x^2(x-1)] = [A(x^2)(x-1)]/[x] + [B(x^2)(x-1)]/[x^2] + [C(x^2)(x-1)]/[x-1]
2x-3 = A(x)(x-1) + B(x-1) + C(x^2)
3 answers