Let's solve the given expression:
[tex]\[
\frac{x+3}{x^2-2x-3} \div \frac{x^2+2x-3}{x+1}
\][/tex]
First, recall that dividing by a fraction is equivalent to multiplying by its reciprocal. Therefore, we can rewrite the division as a multiplication:
[tex]\[
\frac{x+3}{x^2-2x-3} \times \frac{x+1}{x^2+2x-3}
\][/tex]
Next, let's simplify the given fractions by factoring the polynomials in the numerator and denominator:
1. Factor [tex]\(\mathbf{x^2 - 2x - 3}\)[/tex]:
[tex]\[
x^2 - 2x - 3 = (x - 3)(x + 1)
\][/tex]
Hence the expression becomes:
[tex]\[
\frac{x+3}{(x-3)(x+1)}
\][/tex]
2. Factor [tex]\(\mathbf{x^2 + 2x - 3}\)[/tex]:
[tex]\[
x^2 + 2x - 3 = (x + 3)(x - 1)
\][/tex]
Therefore, the reciprocate will be:
[tex]\[
\frac{x+1}{(x+3)(x-1)}
\][/tex]
Thus the given multiplication becomes:
[tex]\[
\frac{x+3}{(x-3)(x+1)} \times \frac{x+1}{(x+3)(x-1)}
\][/tex]
Now, we can multiply the numerators together and the denominators together:
[tex]\[
\frac{(x+3) \cdot (x+1)}{(x-3)(x+1) \cdot (x+3)(x-1)}
\][/tex]
Next, we can cancel out common factors in the numerator and the denominator:
1. [tex]\(x+3\)[/tex] in the numerator and denominator.
2. [tex]\(x+1\)[/tex] in the numerator and denominator.
Leaving us with:
[tex]\[
\frac{1}{(x-3)(x-1)}
\][/tex]
Let's expand the simplified denominator:
[tex]\[
(x-3)(x-1) = x^2 - x - 3x + 3 = x^2 - 4x + 3
\][/tex]
Thus, the simplified expression is:
[tex]\[
\frac{1}{x^2 - 4x + 3}
\][/tex]
From the given choices, the correct answer is:
[tex]\[
\boxed{B}
\][/tex]
