Answer
$\dfrac{x^2+2x+7}{x(x-1)^2}=\dfrac{7}{x}-\dfrac{6}{x-1}+\dfrac{10}{(x-1)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{x^2+2x+7}{x(x-1)^2}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{x^2+2x+7}{x(x-1)^2}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{(x-1)^2}$
Multiply all terms by the least common denominator $x(x-1)^2$:
$x(x-1)^2\cdot\dfrac{x^2+2x+7}{x(x-1)^2}=x(x-1)^2\cdot\dfrac{A}{x}+x(x-1)^2\cdot\dfrac{B}{x-1}+x(x-1)^2\cdot\dfrac{C}{(x-1)^2}$
$x^2+2x+7=A(x-1)^2+Bx(x-1)+Cx$
$x^2+2x+7=Ax^2-2Ax+A+Bx^2-Bx+Cx$
$x^2+2x+7=(A+B)x^2+(-2A-B+C)x+A$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=1\\
-2A-B+C=2\\
A=7
\end{cases}$
Solve the system:
$A=7$
$7+B=1$
$B=-6$
$-2(7)-(-6)+C=2$
$-14+6+C=2$
$-8+C=2$
$C=10$
The partial fraction decomposition is:
$\dfrac{x^2+2x+7}{x(x-1)^2}=\dfrac{7}{x}-\dfrac{6}{x-1}+\dfrac{10}{(x-1)^2}$
