Answer
$\dfrac{x^3-4x^2+9x-5}{(x^2-2x+3)^2}=\dfrac{x-2}{x^2-2x+3}+\dfrac{2x+1}{(x^2-2x+3)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{x^3-4x^2+9x-5}{(x^2-2x+3)^2}$
As the denominator is factored, we can write the partial fraction decomposition:
$\dfrac{x^3-4x^2+9x-5}{(x^2-2x+3)^2}=\dfrac{Ax+B}{x^2-2x+3}+\dfrac{Cx+D}{(x^2-2x+3)^2}$
Multiply all terms by the least common denominator $(x^2-2x+3)^2$:
$(x^2-2x+3)^2\cdot \dfrac{x^3-4x^2+9x-5}{(x^2-2x+3)^2}=(x^2-2x+3)^2\cdot\dfrac{Ax+B}{x^2-2x+3}+(x^2-2x+3)^2\cdot\dfrac{Cx+D}{(x^2-2x+3)^2}$
$x^3-4x^2+9x-5=(Ax+B)(x^2-2x+3)+(Cx+D)$
$x^3-4x^2+9x-5=Ax^3-2Ax^2+3Ax+Bx^2-2Bx+3B+Cx+D$
$x^3-4x^2+9x-5=Ax^3+(-2A+B)x^2+(3A-2B+C)x+(3B+D)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A=1\\
-2A+B=-4\\
3A-2B+C=9\\
3B+D=-5
\end{cases}$
Solve the system:
$A=1$
$\begin{cases}
-2(1)+B=-4\\
3(1)-2B+C=9\\
3B+D=-5
\end{cases}$
$\begin{cases}
-2+B=-4\\
3-2B+C=9\\
3B+D=-5
\end{cases}$
$B=-2$
$3-2B+C=9$
$3-2(-2)+C=9$
$7+C=9$
$C=2$
$3B+D=-5$
$3(-2)+D=-5$
$-6+D=-5$
$D=1$
The partial fraction decomposition is:
$\dfrac{x^3-4x^2+9x-5}{(x^2-2x+3)^2}=\dfrac{x-2}{x^2-2x+3}+\dfrac{2x+1}{(x^2-2x+3)^2}$
