How to convert this polynomial to partial fraction?

Question

I want to convert this polynomoial to partial fraction.

$$ \frac{x^2-2x+2}{x(x-1)} $$

I proceed like this: $$ \frac{x^2-2x+2}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1} $$ Solving, $$ A=-2,B=1 $$ But this does not make sense. What is going wrong?

Answer 1

As the highest power of $x$ in the numerator & the denominator are same,

using Partial Fraction Decomposition, $$\frac{x^2-2x+2}{x(x-1)}=1+\frac Ax+\frac B{x-1}$$

$1$ is found by $$\frac{\text{The coefficient of the highest power of }x\text{ in the numerator}}{\text{The coefficient of the highest power of }x\text{ in the denominator}}$$

Answer 2

$$\frac{x^2-2x+2}{x(x-1)}=1+\frac Ax+\frac B{x-1}$$

Now what you have to do to solve for A and B is to multiply both sides of your equation by $x(x-1)$, and that should give you something like.

$$\frac{x^2-2x+2}{1}=x^2-x +A(x-1)+Bx=x(A+B)-A=-x+2$$ From here on it's pretty easy to solve for both A and B.

Answer 3

An idea I didn't see in the other answers:

$$\frac{x^2-2x+2}{x(x-1)}=\frac{(x-1)^2+1}{x(x-1)}=\frac{x-1}x+\frac1{x(x-1)}=1-\frac1x+\frac1{x(x-1)}$$

And now either directly: $\;\frac1{x(x-1)}=\frac1{x-1}-\frac1x\;$ ,or by partial fractions, so that finally

$$\frac{x^2-2x+2}{x(x-1)}=1-\frac2x+\frac1{x-1}$$