This is a problem I once encountered with a student while working as an online tutor.
\[\int\frac{x^2dx}{(3+4x-4x^2)^{3/2}}\]The first step is to complete the square on the bottom because then we can use some nice trigonometric identities like \(1-\sin^2x=\cos^2x\).
\[3+4x-4x^2=-(4x^2-4x)+3=-(4x^2-4x+1)+4=4-(2x-1)^2\]Now we would like to have \(2x-1=2\sin\t\) so that we can apply a nice trigonometric identity on the bottom. So we use this for the substitution, which is also written as \(x=\frac{1}{2}(2\sin\t+1)\).
\[\int\frac{x^2dx}{(4-(2x-1)^2)^{3/2}} =\int\frac{\left(\frac{1}{2}(2\sin\t+1)\right)^2d(\frac{1}{2}(2\sin\t+1))} {(4-4\sin^2\t)^{3/2}}\]Take out each \(\frac{1}{2}\) on the numerator to simplify it a little.
\[\frac{1}{4}\int\frac{(2\sin\t+1)^2\cos\t d\t}{(4-4\sin^2\t)^{3/2}}\]Now on the bottom, \(4-4\sin^2\t=4\cos^2\t\) and the \(3/2\) exponent turns this into \(8\cos^3\t\).
\[\frac{1}{4}\int\frac{(2\sin\t+1)^2\cos\t d\t}{8\cos^3\t} =\frac{1}{32}\int\frac{(2\sin\t+1)^2d\t}{\cos^2\t} =\frac{1}{32}\int\frac{(4\sin^2\t+4\sin\t+1)d\t}{\cos^2\t}\]Now we will use \(\sin^2\t=1-\cos^2\t\) so part of this integral becomes integrating a constant.
\[\begin{align}&\frac{1}{32}\int\frac{(4-4\cos^2\t+4\sin\t+1)d\t}{\cos^2\t} =\frac{1}{32}\int\frac{(-4\cos^2\t+4\sin\t+5)d\t}{\cos^2\t}\\ &=\frac{1}{32}\int\left(-4+\frac{4\sin\t}{\cos^2\t}+5\sec^2\t\right)d\t =\frac{1}{32}\left(-4\t+\int\frac{4\sin\t d\t}{\cos^2\t}+5\tan\t\right)\\ &=\frac{1}{32}\left(-4\t-\int\frac{4d(\cos\t)}{\cos^2\t}+5\tan\t\right) =\frac{1}{32}\left(-4\t+\frac{4}{\cos\t}+5\tan\t\right)+C\end{align}\]Next, we need to find how to replace all the \(\t\) stuff with \(x\). Our substitution gives us \(\sin\t=\frac{1}{2}(2x-1)\) so we can draw a right triangle with this angle and 2 side lengths, solving for the remaining side length with the Pythagorean theorem.
From this, we have \(\cos\t=\frac{\sqrt{3-4x-4x^2}}{2}\) and \(\tan\t=\frac{2x-1}{\sqrt{3+4x-4x^2}}\). Also using the property that \(\arcsin\) is odd.
\[\begin{align}& \frac{1}{32}\left(-4\arcsin\left(x-\frac{1}{2}\right)+\frac{8}{\sqrt{3+4x-4x^2}} +\frac{5(2x-1)}{\sqrt{3+4x-4x^2}}\right)+C\\ &=\frac{4\arcsin\left(\frac{1}{2}-x\right)\sqrt{3+4x-4x^2}+10x+3} {32\sqrt{3+4x-4x^2}}+C \end{align}\]