Answer
$\displaystyle \frac{x-3}{x-1}$ , $\; x\neq-7$
Work Step by Step
When one denominator is the opposite, or additive inverse of the other,
first multiply either rational expression by $\displaystyle \frac{-1}{-1}$
to obtain a common denominator.
$\displaystyle \frac{x^{2}-2}{x^{2}+6x-7}+\frac{19-4x}{7-6x-x^{2}} \cdot \displaystyle \frac{-1}{-1}= \displaystyle \frac{x^{2}-2}{x^{2}+6x-7}+\frac{-(19-4x)}{-(7-6x-x^{2})}$
$= \displaystyle \frac{x^{2}-2}{x^{2}+6x-7}+\frac{4x-19}{x^{2}+6x-7} $
... To add/subtract rational expressions with the same denominator,
add/subtract numerators and place the sum/difference over the common denominator.
$= \displaystyle \frac{x^{2}+4x-21}{x^{2}+6x-7}$
... factor both trinomials $x^{2}+bx+c$.
... find factors of $c$ whose sum is $b$
$=\displaystyle \frac{(x+7)(x-3)}{(x+7)(x-1)}$
... common factors cancel,
= $\displaystyle \frac{x-3}{x-1}$ , $\; x\neq-7$
