Answer
$$\displaystyle{\int\frac{x^2}{\left(3+4x-4x^2\right)^\frac{3}{2}}dx=\frac{5\left(2x-1\right)+8}{32{\sqrt{3+4x-4x^2}}}-\frac{1}{8}\arcsin\left(\frac{2x-1}{2}\right)+C}$$
Work Step by Step
$\displaystyle{I=\int\frac{x^2}{\left(3+4x-4x^2\right)^\frac{3}{2}}dx}\\ \displaystyle{I=\int\frac{x^2}{\left(4-\left(2x-1\right)^2\right)^\frac{3}{2}}dx}\\$
$\displaystyle \left[\begin{array}{ll} 2x-1=2\sin\theta & \left(2x-1\right)^2=4\sin^2\theta \\ & \\ \frac{dx}{d\theta}=\cos\theta & dx=\cos\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int\frac{\left(\sin\theta+\frac{1}{2}\right)^2}{\left(4-4\sin^2\theta\right)^\frac{3}{2}}\cos\theta\ d\theta}\\ \displaystyle{I=\int\frac{\left(\sin\theta+\frac{1}{2}\right)^2}{\left(4\left(1-\sin^2\theta\right)\right)^\frac{3}{2}}\cos\theta\ d\theta}\\ \displaystyle{I=\int\frac{\sin^2\theta+\sin\theta+\frac{1}{4}}{8\cos^3\theta}\cos\theta\ d\theta}\\ \displaystyle{I=\frac{1}{8}\int\frac{\sin^2\theta+\sin\theta+\frac{1}{4}}{\cos^2\theta}\ d\theta}\\ \displaystyle{I=\frac{1}{8}\int\frac{\sin^2\theta}{\cos^2\theta}+\frac{\sin\theta}{\cos^2\theta}+\frac{1}{4\cos^2\theta}\ d\theta}\\ \displaystyle{I=\frac{1}{8}\int\tan^2\theta+\tan\theta\sec\theta+\frac{1}{4}\sec^2\theta\ d\theta}\\ \displaystyle{I=\frac{1}{8}\int\tan^2\theta+\tan\theta\sec\theta+\frac{1}{4}\sec^2\theta\ d\theta}\\ \displaystyle{I=\frac{1}{8}\int\sec^2\theta-1+\tan\theta\sec\theta+\frac{1}{4}\sec^2\theta\ d\theta}\\ \displaystyle{I=\frac{1}{8}\int\frac{5}{4}\sec^2\theta+\tan\theta\sec\theta-1\ d\theta}\\ \displaystyle{I=\frac{1}{8}\left(\frac{5}{4}\tan\theta+\sec\theta-\theta\right)+C}\\$
$\tan\theta=\frac{2x-1}{\sqrt{3+4x-4x^2}}\\ \sec\theta=\frac{2}{\sqrt{3+4x-4x^2}}\\ $
$\displaystyle{I=\frac{5\left(2x-1\right)}{32{\sqrt{3+4x-4x^2}}}+\frac{2}{8{\sqrt{3+4x-4x^2}}}-\frac{1}{8}\arcsin\left(\frac{2x-1}{2}\right)+C}\\ \displaystyle{I=\frac{5\left(2x-1\right)+8}{32{\sqrt{3+4x-4x^2}}}-\frac{1}{8}\arcsin\left(\frac{2x-1}{2}\right)+C}\\ $
