Answer
$$\frac{2x^2+x-3}{\left(x+3\right)^2\left(x-3\right)}$$
Work Step by Step
Simplifying the expression by creating like denominators, we find:
$$\frac{x}{\left(x+3\right)\left(x-3\right)}+\frac{x+1}{\left(x+3\right)^2} \\ \frac{x\left(x+3\right)}{\left(x+3\right)^2\left(x-3\right)}+\frac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)} \\ \frac{x\left(x+3\right)+\left(x+1\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)} \\ \frac{2x^2+x-3}{\left(x+3\right)^2\left(x-3\right)}$$
