Answer
The partial fraction is, $\frac{1}{x}+\frac{2}{\left( x+7 \right)}-\frac{28}{{{\left( x+7 \right)}^{2}}}$
Work Step by Step
$\frac{3{{x}^{2}}+49}{x{{\left( x+7 \right)}^{2}}}=\frac{A}{x}+\frac{B}{\left( x+7 \right)}+\frac{C}{{{\left( x+7 \right)}^{2}}}$
Now, multiply both sides by $x{{\left( x+7 \right)}^{2}}$; then we get,
$\begin{align}
& 3{{x}^{2}}+49=A{{\left( x+7 \right)}^{2}}+Bx\left( x+7 \right)+Cx \\
& =A\left( {{x}^{2}}+49+14x \right)+B{{x}^{2}}+7Bx+Cx \\
& =A{{x}^{2}}+49A+14Ax+B{{x}^{2}}+7Bx+Cx \\
& =\left( A+B \right){{x}^{2}}+\left( 14A+7B+C \right)x+49A
\end{align}$
Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term:
$A+B=3$ …… (1)
$14A+7B+C=0$ …… (2)
$49A=49$ …… (3)
Since, $49A=49$. It implies:
$A=1$
Put $A=1$ in equation (1) to get:
$\begin{align}
& 1+B=3 \\
& B=3-1 \\
& =2
\end{align}$
It implies $B=2$
Now, put $A=1$ and $B=2$ in equation (2); then we get:
$\begin{align}
& 14\left( 1 \right)+7\left( 2 \right)+C=0 \\
& 28+C=0 \\
& C=-28
\end{align}$
Therefore,
$\frac{3{{x}^{2}}+49}{x{{\left( x+7 \right)}^{2}}}=\frac{1}{x}+\frac{2}{\left( x+7 \right)}-\frac{28}{{{\left( x+7 \right)}^{2}}}$
Thus, the partial fraction of the provided expression is $\frac{1}{x}+\frac{2}{\left( x+7 \right)}-\frac{28}{{{\left( x+7 \right)}^{2}}}$.
