Partial Fractions of 4/(x^3-2x^2)

[MasterJan7]

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!

 

[LCKurtz]

I assume you know yours is incorrect because it doesn't expand to equal the given fraction.

The correct expansion is
$$\frac{4}{x^2(x-2)}=\frac{Ax+B}{x^2}+\frac{C}{x-2}$$

 

[SammyS]

You should have something like: $\displaystyle \frac{4}{x^2(x-2)}=\frac{Ax+B}{x^2}+\frac{C}{x-2}\frac{}{}$

or equivalently: $\displaystyle \frac{4}{x^2(x-2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-2}\frac{}{}$