Answer
$\dfrac{(x+1)(x+2)}{(x-1)(x-2)} $
Work Step by Step
We apply Method 1 (A51), to treat the numerator and denominator of the complex rational expression separately. So by factoring both the denominator and numerator of each rational function and cancelling the common factors, we have
$$
\frac{\frac{x^2+7x+6}{x^2+x-6}}{\frac{x^2+5x-6}{x^2+5x+6}}=\frac{\frac{(x+1)(x+6)}{(x+3)(x-2)}}{\frac{(x-1)(x+6)}{(x+3)(x+2)}}=\frac{(x+1)(x+6)}{(x+3)(x-2)}\frac{(x+3)(x+2)}{(x-1)(x+6)}\\
=\frac{(x+1)(x+2)}{(x-1)(x-2)}
.$$
