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## Homework Statement

Find partial fractions for 4/(x^3-2x^2)

## The Attempt at a Solution

Heres the steps that I took:

1. 4/(x^3-2x^2)= 4/(x^2(x-2))= A/(x^2) + B/(x-2)

2. 4= A(x-2) + B(x^2)

3. When x=0, -2A=4, so A=-2,

and When x=2, 4B=4, so B=1.

4. So my final answer was:

-2/(x^2)+1/(x-2)

The real answer as I found out from Wolfram Alpha integral calculator was:

-2/(x^2)+1/(x-2)-1/x

So the real answer is the same as the solution that I got, except for the -1/x at the end... I have no idea where that -1/x came from, no matter how many times I redo this problem. Please tell how to get the real answer! Thank you!