Answer
$=\displaystyle \frac{2x-1}{x+3}, \quad x\neq 0,-3$
Work Step by Step
The fractions have a common denominator, we add the numerators...
$\displaystyle \frac{x^{2}-2x}{x^{2}+3x}+\frac{x^{2}+x}{x^{2}+3x}=\frac{x^{2}-2x+x^{2}+x}{x^{2}+3x}$
$=\displaystyle \frac{2x^{2}-x}{x^{2}+3x}$
... factor what we can...
$=\displaystyle \frac{x(2x-1)}{x(x+3)}$
... exclude values that yield 0 in the denominator:
$x\neq 0,-3$
... and, cancel the common factor
$=\displaystyle \frac{2x-1}{x+3}, \quad x\neq 0,-3$
