Answer
$\dfrac{x^4-x^2+2}{x^3-x^2}=x+1-\dfrac{2}{x}-\dfrac{2}{x^2}+\dfrac{2}{x-1}$
Work Step by Step
We are given the fraction:
$\dfrac{x^4-x^2+2}{x^3-x^2}$
Because the degree of the numerator is greater than the degree of the denominator, we divide the numerator by the denominator:
$\dfrac{x^4-x^2+2}{x^3-x^2}=\dfrac{x(x^3-x^2)+x^3-x^2+2}{x^3-x^2}=\dfrac{x(x^3-x^2)+(x^3-x^2)+2}{x^3-x^2}=\dfrac{(x^3-x^2)(x+1)+2}{x^3-x^2}=x+1+\dfrac{2}{x^3-x^2}$
We will write the partial fraction decomposition for the fraction $\dfrac{2}{x^3-x^2}$. Factor the denominator and write the partial fraction decomposition:
$\dfrac{2}{x^3-x^2}=\dfrac{2}{x^2(x-1)}$
Write the partial fraction decomposition:
$\dfrac{2}{x^3-x^2}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x-1}$
Multiply all terms by the least common denominator $x^2(x-1)$:
$x^2(x-1)\cdot\dfrac{2}{x^3-x^2}=x^2(x-1)\cdot\dfrac{A}{x}+x^2(x-1)\cdot\dfrac{B}{x^2}+x^2(x-1)\cdot\dfrac{C}{x-1}$
$2=Ax(x-1)+B(x-1)+Cx^2$
$2=Ax^2-Ax+Bx-B+Cx^2$
$2=(A+C)x^2+(-A+B)x-B$
Identify the coefficients:
$\begin{cases}
A+C=0\\
-A+B=0\\
-B=2
\end{cases}$
$B=-2$
$-A+B=0$
$-A+(-2)=0$
$A=-2$
$A+C=0$
$-2+C=0$
$C=2$
The partial fraction decomposition is:
$\dfrac{2}{x^2(x-1)}=-\dfrac{2}{x}-\dfrac{2}{x^2}+\dfrac{2}{x-1}$
The final result is:
$\dfrac{x^4-x^2+2}{x^3-x^2}=x+1-\dfrac{2}{x}-\dfrac{2}{x^2}+\dfrac{2}{x-1}$
