I have absolutely no idea how to approach this question:
$$\int \frac{x^2}{(15+6x-9x^2)^{3/2}} \ \mathrm{d}x$$
I'm almost positive that it has something to do with trigonometric substitution, but no matter how much I rearrange the equation or try to perform substitution I'm just not seeing how to solve it.
To prepare for the trigonometric substitution, we complete the square, noting that $15+6x-9x^2=16-(3x-1)^2$. Let $3x-1=4\sin\theta$. Then $3\,dx=4\cos\theta\,d\theta$, and our integral can be rewritten as $$\int \frac{(1+4\sin \theta)^2}{9}\frac{1}{8\cos^3 \theta}\frac{4}{3}\cos\theta\,d\theta.$$
Our integral then simplifies to $$\frac{1}{54}\int\left(\sec^2\theta+8\sec\theta\tan\theta+16\tan^2\theta \right)\,d\theta.$$ Now we are essentially finished, for $\int\sec^2\theta\,d\theta=\tan\theta+C$, while $\int \sec\theta\tan\theta=\sec\theta+C$. Finally, to integrate $\tan^2\theta$ rewrite as $\sec^2\theta-1$.
To express the integral in terms of $x$, draw as usual a right triangle with hypotenuse $4$ and leg $3x-1$ opposite $\angle\theta$. Then we csn use the Pythagorean Theorem to write down the length of the other leg, and read off the trigonometric functions of $\theta$ from the picture.