Integrate the function: $\frac{1}{\sqrt{8+3x-x^{2}}}$

To integrate $\int \frac{1}{\sqrt{8+3x-x^{2}}} dx$,we first complete the square for the quadratic expression $8+3x-x^{2}$.
$8+3x-x^{2} = - (x^{2}-3x-8)$
$= - \left(x^{2}-3x + (\frac{3}{2})^{2} - (\frac{3}{2})^{2} - 8\right)$
$= - \left((x-\frac{3}{2})^{2} - \frac{9}{4} - 8\right)$
$= - \left((x-\frac{3}{2})^{2} - \frac{41}{4}\right)$
$= \frac{41}{4} - (x-\frac{3}{2})^{2}$
Now,the integral becomes $\int \frac{1}{\sqrt{\frac{41}{4} - (x-\frac{3}{2})^{2}}} dx$.
Let $t = x-\frac{3}{2}$,then $dt = dx$.
Using the formula $\int \frac{1}{\sqrt{a^{2}-t^{2}}} dt = \sin^{-1}(\frac{t}{a}) + C$,where $a = \frac{\sqrt{41}}{2}$:
$= \sin^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right) + C$
$= \sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right) + C$,where $C$ is an arbitrary constant.