Question:
$\sqrt{1+3 x-x^{2}}$
Solution:
Let $I=\int \sqrt{1+3 x-x^{2}} d x$
$=\int \sqrt{1-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)} d x$
$=\int \sqrt{\left(1+\frac{9}{4}\right)-\left(x-\frac{3}{2}\right)^{2}} d x$
$=\int \sqrt{\left(\frac{\sqrt{13}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}} d x$
It is known that, $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+\mathrm{C}$
$\begin{aligned} \therefore I &=\frac{x-\frac{3}{2}}{2} \sqrt{1+3 x-x^{2}}+\frac{13}{4 \times 2} \sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}}\right)+\mathrm{C} \\ &=\frac{2 x-3}{4} \sqrt{1+3 x-x^{2}}+\frac{13}{8} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{13}}\right)+\mathrm{C} \end{aligned}$