Answer
$\{-1\}$.
Work Step by Step
First we determine the Least Common Denominator (LCD) for all the denominators in the given expression in order to clear fractions.
Factor $x^2+2x-3$.
Rewrite the middle term $2x$ as $3x-1x$.
$\Rightarrow x^2+3x-1x-3$
Group the terms.
$\Rightarrow (x^2+3x)+(-1x-3)$
Factor each group.
$\Rightarrow x(x+3)-1(x+3)$
Factor out $x+3$.
$\Rightarrow (x+3)(x-1)$
Back substitute the factor into the given equation.
$\Rightarrow \frac{x+1}{(x+3)(x-1)}-\frac{1}{x+3}=\frac{1}{x-1}$
The LCD of the fractions is $(x+3)(x-1)$.
Multiply the equation by LCD.
$\Rightarrow (x+3)(x-1)\left (\frac{x+1}{(x+3)(x-1)}-\frac{1}{x+3}\right )=(x+3)(x-1)\left (\frac{1}{x-1}\right)$
Use distributive property.
$\Rightarrow (x+3)(x-1)\left (\frac{x+1}{(x+3)(x-1)}\right )-(x+3)(x-1)\left (\frac{1}{x+3}\right )=(x+3)(x-1)\left (\frac{1}{x-1}\right)$
Cancel common factors.
$\Rightarrow x+1-x+1=x+3$
Add like terms.
$\Rightarrow 2=x+3$
Subtract $3$ from both sides.
$\Rightarrow 2-3=x+3-3$
Simplify.
$\Rightarrow -1=x$
The solution set is $\{-1\}$.
Note: The given equation is defined for all real values of $x$ except the zeros of the denominators, which are $-3$ and $1$. Because our solution ($x=-1$) is not a zero of the denominator, it follows that it is the solution is correct.
