Answer
$$\frac{2a^2+5a+2}{a^2-4a+3}$$
Work Step by Step
Recall, one never divides by a fraction. Rather, if we want to divide by a fraction, we multiply by the reciprocal of the fraction. Recall, the reciprocal of a fraction is that fraction with the numerator and the denominator switched. Thus, we find:
$$ \frac{a^2-a-6}{a^2-1}\times \frac{2a^2+3a+1}{a^2-6a+9}\\ \frac{\left(a+2\right)\left(a-3\right)\left(2a+1\right)\left(a+1\right)}{\left(a^2-1\right)\left(a^2-6a+9\right)}\\ \frac{\left(a+2\right)\left(2a+1\right)\left(a+1\right)}{\left(a+1\right)\left(a-1\right)\left(a-3\right)}\\ \frac{\left(a+2\right)\left(2a+1\right)}{\left(a-1\right)\left(a-3\right)}\\ \frac{2a^2+5a+2}{a^2-4a+3}$$
