Answer
The partial fraction is, $\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$
Work Step by Step
$\frac{{{x}^{2}}+2x+7}{x{{\left( x-1 \right)}^{2}}}=\frac{A}{x}+\frac{B}{\left( x-1 \right)}+\frac{C}{{{\left( x-1 \right)}^{2}}}$
Now, multiply both sides by
$\begin{align}
& {{x}^{2}}+2x+7=A{{\left( x-1 \right)}^{2}}+Bx\left( x-1 \right)+Cx \\
& =A\left( {{x}^{2}}+1-2x \right)+B{{x}^{2}}-Bx+Cx \\
& =A{{x}^{2}}+A-2Ax+B{{x}^{2}}-Bx+Cx \\
& =\left( A+B \right){{x}^{2}}+\left( -2A-B+C \right)x+A
\end{align}$
Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term:
$A+B=1$ …… (1)
$-2A-B+C=2$ …… (2)
$A=7$ …… (3)
And put the value of A into equation (1):
$\begin{align}
& 7+B=1 \\
& B=-6
\end{align}$
Also, put the value of A and B into equation (2):
$\begin{align}
& -2\left( 7 \right)+6+C=2 \\
& C=2+8 \\
& =10
\end{align}$
Therefore,
$\frac{{{x}^{2}}+2x+7}{x{{\left( x-1 \right)}^{2}}}=\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$
Thus, the partial fraction of the provided expression is $\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$.
