$\int \dfrac {2x + 5}{\sqrt {7 - 6x - x^{2}}} dx = A\sqrt {7 - 6x - x^{2}} + B\sin^{-1} \left (\dfrac {x + 3}{4}\right ) + C$

(where $C$ is a constant of integration), then the ordered pair $(A, B)$ is equal to

Note that $7-6x-x^2 = 16 - (x+3)^2$ and $\dfrac{d}{dx} (7-6x-x^2) = -2x-6$

So, we have

$\int \dfrac{2x+5}{\sqrt{7-6x-x^2}} dx = \int \dfrac{2x+6}{\sqrt{7-6x-x^2}} dx - \int \dfrac{1}{\sqrt{16-(x+3)^2}} dx\\ = -2\sqrt{7-6x-x^2} - \sin^{-1} (\dfrac{x+3}{4}) + C$

So, we have $A = -2, \ B = -1$.