#### Answer

$\displaystyle \frac{x-3}{x+2}, \qquad x\neq-1,\ 3,\ -2$

#### Work Step by Step

Factor all denominators of the complex rational expression.
(For $x^{2}-2x-3$, we find factors of $-3$ that add to $-2$.
These are $+1$ and $-3.$)
$\displaystyle \frac{\frac{1}{x+1}}{\frac{1}{x^{2}-2x-3}+\frac{1}{x-3}}=\frac{\frac{1}{x+1}}{\frac{1}{(x+1)(x-3)}+\frac{1}{x-3}}$
Complex rational expressions have rational expressions in the numerator or/and in the denominator.
Here, multiplying both the numerator and denominator with $(x+1)(x-3)$ would get rid of these fractions.
We have to exclude any value that yields 0 in any denominator.
Excluded: $x=-1,3$.
The denominator of the expression itself can not be zero,
so we will exclude any values for which it is 0:
$\displaystyle \frac{1}{(x+1)(x-3)}+\frac{1}{x-3}=0\qquad/\times(x+1)(x-3)$
$1+1(x+1)=0$
$1+x+1=$
$x+2=0$
Excluded: $x=-1,\ 3,\ -2$
$=\displaystyle \frac{\frac{1}{x+1}}{\frac{1}{(x+1)(x-3)}+\frac{1}{x-3}}\times\frac{(x+1)(x-3)}{(x+1)(x-3)}$
... the denominator was simplified while solving the above equation.
$=\displaystyle \frac{x-3}{x+2}, \qquad x\neq-1,\ 3,\ -2$