Integrate the function: $\frac{1}{\sqrt{7-6x-x^{2}}}$

(N/A) To integrate $\int \frac{1}{\sqrt{7-6x-x^{2}}} dx$,we first complete the square for the quadratic expression $7-6x-x^{2}$.
$7-6x-x^{2} = 7 - (x^{2} + 6x)$
$= 7 - (x^{2} + 6x + 9 - 9)$
$= 7 - ((x+3)^{2} - 9)$
$= 7 + 9 - (x+3)^{2}$
$= 16 - (x+3)^{2}$
$= (4)^{2} - (x+3)^{2}$
Now,the integral becomes:
$\int \frac{1}{\sqrt{(4)^{2} - (x+3)^{2}}} dx$
Let $u = x+3$,then $du = dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^{2}-u^{2}}} du = \sin^{-1}(\frac{u}{a}) + C$,we get:
$= \sin^{-1}(\frac{x+3}{4}) + C$,where $C$ is the constant of integration.