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

We have to evaluate the integral

$$\int \dfrac{2x+5}{\sqrt{7-6x-x^{2}}}\,dx$$

and then compare it with the given expression

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

First we simplify the square-root in the denominator by completing the square. Write

$$7-6x-x^{2}=-(x^{2}+6x-7).$$

Inside the brackets we add and subtract the square of half the coefficient of $$x$$:

$$x^{2}+6x-7=x^{2}+6x+9-9-7=(x+3)^{2}-16.$$

Therefore

$$7-6x-x^{2}=-(x+3)^{2}+16=16-(x+3)^{2},$$

so that

$$\sqrt{7-6x-x^{2}}=\sqrt{16-(x+3)^{2}}.$$

Substituting this result into the integral gives

$$\int \dfrac{2x+5}{\sqrt{16-(x+3)^{2}}}\,dx.$$

Now we set

$$u=x+3\quad\Longrightarrow\quad du=dx,\qquad x=u-3.$$

With this change of variable the numerator becomes

$$2x+5=2(u-3)+5=2u-6+5=2u-1,$$

and the integral takes the form

$$I=\int \dfrac{2u-1}{\sqrt{16-u^{2}}}\,du.$$

We split the integrand into two simpler parts:

$$I=\int \dfrac{2u}{\sqrt{16-u^{2}}}\,du-\int \dfrac{1}{\sqrt{16-u^{2}}}\,du.$$

We evaluate each integral separately.

For the first integral we recognise the standard form $$\int \dfrac{u}{\sqrt{a^{2}-u^{2}}}\,du=-\sqrt{a^{2}-u^{2}}+C.$$ Taking $$a=4$$ (because $$a^{2}=16$$) we write

$$\int \dfrac{2u}{\sqrt{16-u^{2}}}\,du =2\int \dfrac{u}{\sqrt{16-u^{2}}}\,du =2\left[-\sqrt{16-u^{2}}\right] =-2\sqrt{16-u^{2}}.$$

For the second integral we use the standard formula $$\int \dfrac{du}{\sqrt{a^{2}-u^{2}}}=\sin^{-1}\!\left(\dfrac{u}{a}\right)+C.$$ Again $$a=4$$, so

$$\int \dfrac{1}{\sqrt{16-u^{2}}}\,du =\sin^{-1}\!\left(\dfrac{u}{4}\right).$$

Combining the two results we obtain

$$I=-2\sqrt{16-u^{2}}-\sin^{-1}\!\left(\dfrac{u}{4}\right)+C.$$

Now we reverse the substitution $$u=x+3$$:

$$I=-2\sqrt{16-(x+3)^{2}}-\sin^{-1}\!\left(\dfrac{x+3}{4}\right)+C.$$

But earlier we found $$\sqrt{16-(x+3)^{2}}=\sqrt{7-6x-x^{2}}$$, so

$$I=-2\sqrt{7-6x-x^{2}}-\sin^{-1}\!\left(\dfrac{x+3}{4}\right)+C.$$

Comparing this with the desired form

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

we read off

$$A=-2,\qquad B=-1.$$

Hence, the ordered pair $$(A,B)$$ equals $$(-2,-1)$$.

Hence, the correct answer is Option A.