Question

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

Answer 1

$\int \sqrt{1+3x-x^2}dx$

$=\int \sqrt{1-(x^2-3x)}dx$

$=\int \sqrt{1-[x^2-3x+{\left(\frac{3}{2}\right)}^2-{\left(\frac{3}{2}\right)}^2]}dx$

$=\int \sqrt{1-[{(x-\frac{3}{2})}^2-{\left(\frac{3}{2}\right)}^2]}dx$

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

$=\int \sqrt{\frac{13}{4}-{(x-\frac{3}{2})}^2}dx$

$=\int \sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{(x-\frac{3}{2})}^2}dx$

Using $\int \sqrt{a^2-x^2}.dx$

$=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$

$=\frac{x-\frac{3}{2}}{2}\sqrt{{\left(\frac{\sqrt{13}}{2}\right)}^2-{(x-\frac{3}{2})}^2}+\frac{{\left(\frac{\sqrt{13}}{2}\right)}^2}{2}{\sin }^{-1}\frac{(x-\frac{3}{2})}{\frac{\sqrt{13}}{2}}+C$

$=\frac{\frac{2x-3}{2}}{2}\sqrt{\frac{13}{4}-{(x-\frac{3}{2})}^2}+\frac{\frac{13}{4}}{2}{\sin }^{-1}\frac{\left(\frac{2x-3}{2}\right)}{\frac{\sqrt{13}}{2}}+C$

$=\frac{2x-3}{4}\sqrt{\frac{13}{4}-[x^2+\frac{9}{4}-3x]}+\frac{13}{8}{\sin }^{-1}\frac{2x-3}{\sqrt{13}}+C$

$=\frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}{\sin }^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C$

Where $C$ is constant of integration