<s>[INST] Evaluate integral $\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$ Evaluate the integral
$$\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$$
Edit:
Here is Quanto's brilliant method of doing this problem with the details filled out:
SOLUTION:
First, lets complete the square $-9x^2+6x+15$:
$-9x^2+6x+15$
$=-9(x^2-\frac{2}{3}x)+15$
$=-9(x^2-\frac{2}{3}x+(\frac{2}{6})^2)+15-(-9(\frac{2}{6})^2)$
$=-9(x-\frac{1}{3})^2+16$...
$=-(1-3x)^2+16$
Now we are going to do a few extra steps in completing the square then usual, so this will be easier to integrate.
$=(-(\frac{1-3x}{4})^2+1)16$
Thus our original integral becomes:
$\int \frac{x^2}{(-(\frac{1-3x}{4})^2+1)16)^{\frac{3}{2}}}dx$
Now set $\sin(t) = \frac{1-3x}{4}$. Thus $x=\frac{1-4\sin(t)}{3}$ and $dx = \frac{-4\cos(t)}{3} \, du$
Note the strategic choice of defining $\sin(t) = \frac{1-3x}{4}$, it has a perfect place to plug in in the denominator. Let's make these substitutions
$=\int \frac{(\frac{1-4\sin(t)}{3})^2}{((-\sin^2(t)+1)16)^{\frac{3}{2}}}(\frac{-4\cos(u)}{3} \, du)$
$=\frac{-1}{432}\int \frac{(1-4\sin(t))^2}{(-\sin^2(t)+1)^{\frac{3}{2}}}(\cos(u) \, du)$
$=\frac{-1}{432}\int \frac{(1-4\sin(t))^2}{\cos^2(t)} \, du)$
$=\frac{-1}{432}\int \sec^2(t)+16\tan^2(t)-\frac{8\sin(t)}{\cos^2(t)}$
Now, Applying the identity $\tan^2(x)=\sec^2(x)-1$ we arrive at:
$=\frac{-1}{432}\int 17\sec^2(t)-16-\frac{8\sin(t)}{\cos^2(t)} \, dt$
Let's evaluate each of these separately:
$\int 17\sec^2(t)\,dt = 17\tan(t)$
$\int 16 \, dt = 16t$
$\int \frac{8\sin(t)}{\cos^2(t)}$ can be solved with u-substitution after setting $u=\cos(t)$. to get:
$\int \frac{8\sin(t)}{\cos^2(t)}=\frac{8}{\cos(t)}$
Thus our answer is:
$=\frac{1}{432}(16t-17\tan(t)+\frac{8}{\cos(t)})+C$
 [/INST] Let $\sin t = \frac{1-3x}4$. Then,
\begin{align}
\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}}dx
&=-\frac1{432}\int \frac{(1-4\sin t)^2}{\cos^2t}dt\\
&=\frac1{432}\int\left( 16-17\sec^2 t +\frac{8\sin t}{\cos^2t}\right)dt
\\
&= \frac1{432}\left( 16t -17\tan t+\frac8{\cos t}\right) +C
\end{align}
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