Question

$\int \sqrt{1+3x-x^2}dx$ is equal to
(where $C$ is integration constant)

Answer 1

$I=\int \sqrt{1+3x-x^2}dx=\int \sqrt{1-(x^2-3x+\frac{9}{4}-\frac{9}{4})}dx=\int \sqrt{(1+\frac{9}{4})-{(x-\frac{3}{2})}^2}dx=\int \sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{(x-\frac{3}{2})}^2}dx$
We know that ,
$\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$
$\therefore I=\frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\frac{13}{4\times 2}{\sin }^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}}\right)+C=\frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}{\sin }^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C$