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

Let $I = \int \sqrt{1+3x-x^{2}} dx$
$= \int \sqrt{1 - (x^{2} - 3x + \frac{9}{4} - \frac{9}{4})} dx$
$= \int \sqrt{(1 + \frac{9}{4}) - (x - \frac{3}{2})^{2}} dx$
$= \int \sqrt{(\frac{\sqrt{13}}{2})^{2} - (x - \frac{3}{2})^{2}} dx$
Using the standard integral formula $\int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{x}{a}) + C$,we get:
$I = \frac{x - \frac{3}{2}}{2} \sqrt{1 + 3x - x^{2}} + \frac{13}{4 \times 2} \sin^{-1}\left(\frac{x - \frac{3}{2}}{\frac{\sqrt{13}}{2}}\right) + C$
$= \frac{2x - 3}{4} \sqrt{1 + 3x - x^{2}} + \frac{13}{8} \sin^{-1}\left(\frac{2x - 3}{\sqrt{13}}\right) + C$
Where $C$ is an arbitrary constant.