Prove

$7-6 x-x^{2}$ can be written as $7-\left(x^{2}+6 x+9-9\right)$

Therefore,

$7-\left(x^{2}+6 x+9-9\right)$

$=16-\left(x^{2}+6 x+9\right)$

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

$=(4)^{2}-(x+3)^{2}$

$\therefore \int \frac{1}{\sqrt{7-6 x-x^{2}}} d x=\int \frac{1}{\sqrt{(4)^{2}-(x+3)^{2}}} d x$

Let $x+3=t$

$\Rightarrow d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{(4)^{2}-(x+3)^{2}}} d x=\int \frac{1}{\sqrt{(4)^{2}-(t)^{2}}} d t$

$=\sin ^{-1}\left(\frac{t}{4}\right)+C$

$=\sin ^{-1}\left(\frac{x+3}{4}\right)+C$