f(x) =

1/

(2x − 5)^2(3x − 1)based at b = 0.

I tried partial fractions and I think I'm doing it wrong.

Thanks in advance.

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- Oct 8th 2008, 03:08 PMkhuezyTaylor Series partial fractions
f(x) =

1/

(2x − 5)^2(3x − 1)based at b = 0.

I tried partial fractions and I think I'm doing it wrong.

Thanks in advance. - Oct 8th 2008, 03:40 PMmr fantastic
- Oct 8th 2008, 04:02 PMkhuezyok
1 = A(2x-5)(3x-1) + B(3x-1) + C(2x-5)(2x-5)

x=5/2

1= B(3(5/2)-1) B=2/13????

That doesn't seem right. - Oct 8th 2008, 04:41 PMmr fantastic