Question

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

Answer 1

$$\begin{gathered} \int \sqrt{3+2x-x^2}dx \\ =\int \sqrt{3-\left(x^2-2x\right)}dx \\ =\int \sqrt{3-\left(x^2-2x+1-1\right)}dx \\ =\int \sqrt{4-{\left(x-1\right)}^2}dx \\ =\int \sqrt{2^2-{\left(x-1\right)}^2}dx\left[\because \int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{1}{2}{a}^2{\sin }^{-1}\frac{x}{a}+C\right] \\ =\left(\frac{x-1}{2}\right)\sqrt{2^2-{\left(x-1\right)}^2}+\frac{2^2}{2}{\sin }^{-1}\left(\frac{x-1}{2}\right)+C \\ =\frac{x-1}{2}\sqrt{3+2x-x^2}+{\sin }^{-1}\left(\frac{x-1}{2}\right)+C \end{gathered}$$