Evaluate the following integrals:

$8+3 x-x 2$ can be written as $8-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)$

Therefore

$8-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)$

$=\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}$

$\int \frac{1}{\sqrt{8+3 x-x^{2}}} d x=\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x$

Let $x-3 / 2=t$

$d x=d t$

$\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x=\int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^{2}-t^{2}}} d t$

$=\sin ^{-1}\left(\frac{\mathrm{t}}{\frac{\sqrt{41}}{2}}\right)+\mathrm{c}$

$\left[\right.$ since $\left.\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]$

$=\sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+c$

$=\sin ^{-1}\left(\frac{2 x-3}{\sqrt{41}}\right)+c$