Answer
$$\displaystyle{\int x^2\sqrt{3+2x-x^2}dx=4\arcsin{\left(\frac{x-1}{2}\right)}-\frac{\left(x-1\right){\left(3+2x-x^2\right)^{\frac{3}{2}}}}{8}+\frac{\left(x-1\right)^3{\sqrt{3+2x-x^2}}}{8}+\frac{\left(x-1\right){\sqrt{3+2x-x^2}}}{2}-\frac{2}{3}\left(3+2x-x^2\right)^{\frac{3}{2}}+C}$$
Work Step by Step
$\displaystyle{I=\int x^2\sqrt{3+2x-x^2}dx}\\ \displaystyle{I=\int x^2\sqrt{4-\left(x-1\right)^2}dx}\\$
$\displaystyle \left[\begin{array}{ll} x-1=2\sin\theta & \left(x-1\right)^2=4\sin^2\theta \\ & \\ \frac{dx}{d\theta}=2\cos\theta & dx=2\cos\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int \left(2\sin\theta+1\right)^2\sqrt{4-4\sin^2\theta}\cos\theta\ d\theta}\\ \displaystyle{I=\int \left(2\sin\theta+1\right)^2\sqrt{4\left(1-\sin\theta\right)^2}2\cos\theta\ d\theta}\\ \displaystyle{I=\int \left(2\sin\theta+1\right)^2\left(2\cos\theta\right)2\cos\theta\ d\theta}\\ \displaystyle{I=\int \left(2\sin\theta+1\right)^2\left(2\cos\theta\right)2\cos\theta\ d\theta}\\ \displaystyle{I=\int \left(4\sin^2\theta+4\sin\theta+1\right)\left(4\cos^2\theta\right)\ d\theta}\\ \displaystyle{I=\int16\sin^2\theta\cos^2\theta+4\cos^2\theta+16\sin\theta\cos^2\theta\ d\theta}\\ \displaystyle{I=4\int\left(2\sin\theta\cos\theta\right)^2+\cos^2\theta+4\sin\theta\cos^2\theta\ d\theta}\\ \displaystyle{I=4\int\sin^22\theta+\cos^2\theta+\frac{4}{3}\times3\sin\theta\cos^2\theta\ d\theta}\\ \displaystyle{I=4\int\frac{1}{2}\left(1-\cos4\theta\right)+\frac{1}{2}\left(1+\cos2\theta\right)+\frac{4}{3}\times3\sin\theta\cos^2\theta\ d\theta}\\ \displaystyle{I=2\left(\theta-\frac{1}{4}\sin4\theta\right)+2\left(\theta+\frac{1}{2}\sin2\theta\right)-\frac{16}{3}\cos^3\theta+C}\\ \displaystyle{I=2\theta-\frac{1}{2}\sin4\theta+2\theta+\sin2\theta-\frac{16}{3}\cos^3\theta+C}\\ \displaystyle{I=4\theta-\frac{1}{2}\left(4\sin\theta\cos^3\theta-4\sin^3\theta\cos\theta\right)+2\sin\theta\cos\theta-\frac{16}{3}\cos^3\theta+C}\\ \displaystyle{I=4\theta-2\sin\theta\cos^3\theta+2\sin^3\theta\cos\theta+2\sin\theta\cos\theta-\frac{16}{3}\cos^3\theta+C}\\$
$\sin\theta=\frac{x-1}{2}\\ \cos\theta=\frac{\sqrt{3+2x-x^2}}{2}\\ $
$\displaystyle{I=4\arcsin{\left(\frac{x-1}{2}\right)}-2\left(\frac{x-1}{2}\right)\left(\frac{\left(3+2x-x^2\right)^{\frac{3}{2}}}{8}\right)+2\left(\frac{\left(x-1\right)^3}{8}\right)\left(\frac{\sqrt{3+2x-x^2}}{2}\right)+2\left(\frac{x-1}{2}\right)\left(\frac{\sqrt{3+2x-x^2}}{2}\right)-\frac{16}{3}\left(\frac{\left(3+2x-x^2\right)^{\frac{3}{2}}}{8}\right)+C}\\ \displaystyle{I=4\arcsin{\left(\frac{x-1}{2}\right)}-\frac{\left(x-1\right){\left(3+2x-x^2\right)^{\frac{3}{2}}}}{8}+\frac{\left(x-1\right)^3{\sqrt{3+2x-x^2}}}{8}+\frac{\left(x-1\right){\sqrt{3+2x-x^2}}}{2}-\frac{2}{3}\left(3+2x-x^2\right)^{\frac{3}{2}}+C} $
