$$
\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=A \sqrt{7-6 x-x^2}+B \sin ^{-1}\left(\frac{x+3}{4}\right)+C
$$
(where $C$ is a constant of integration), then the ordered pair $(A, B)$ is equal to
$(-2,-1)$
$(2,-1)$
$(-2,1)$
$(2,1)$
Solution
$\because 7-6 x-x^2=16-(x+3)^2$
and $\frac{d}{d x}\left(7-6 x-x^2\right)=-2 x-6$
So, $\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=\int \frac{2 x+6}{\sqrt{7-6 x-x^2}} d x$
$
-\int \frac{1}{\sqrt{16-(x+3)^2}} d x
$
$
=-2 \sqrt{7-6 x-x^2}-\sin ^{-1}\left(\frac{x+3}{4}\right)+C
$
Therefore, $A=-2, \& B=-1$