Answer
The required value is $\frac{{{x}^{2}}+2x+15}{\left( x+3 \right)\left( x-3 \right)}$ , $x\ne 3,-3$.
Work Step by Step
Consider the expression,
$\frac{x}{x+3}+\frac{5}{x-3}$
The given expression can be further evaluated by taking least common denominator:
$\begin{align}
& \frac{x}{x+3}+\frac{5}{x-3}=\frac{x\left( x-3 \right)}{\left( x+3 \right)\left( x-3 \right)}+\frac{5\left( x+3 \right)}{\left( x+3 \right)\left( x-3 \right)} \\
& =\frac{x\left( x-3 \right)+5\left( x+3 \right)}{\left( x+3 \right)\left( x-3 \right)} \\
& =\frac{{{x}^{2}}-3x+5x+15}{\left( x+3 \right)\left( x-3 \right)} \\
& =\frac{{{x}^{2}}+2x+15}{\left( x+3 \right)\left( x-3 \right)}
\end{align}$
Here, $x\ne 3,-3$ because at $x=3,-3$ the denominator will become zero.
Therefore, the value of $\frac{x}{x+3}+\frac{5}{x-3}$ is $\frac{{{x}^{2}}+2x+15}{\left( x+3 \right)\left( x-3 \right)}$.
