Question

Integrate the function.
$\int \sqrt{1+3x-x^2}dx$

Answer 1

Let $I=\int \sqrt{1+3x-x^2}dx=\int \sqrt{-(x^2-3x-1)}dx$
$=\int \sqrt{-(x^2-3x+(\frac{3}{2}{)}^2-(\frac{3}{2}{)}^2-1)}dx=\int \sqrt{-({(x-\frac{3}{2})}^2-\frac{9}{4}-1)}dx=\int \sqrt{-({(x-\frac{3}{2})}^2-\frac{9+4}{4})}dx=\int \sqrt{-({(x-\frac{3}{2})}^2-\frac{13}{4})}dx=\int \sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{(x-\frac{3}{2})}^2}dx\therefore I=\frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\frac{13}{4\times 2}si{n}^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}}\right)[\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}+C]\Rightarrow I=\frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}si{n}^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C$