<s>[INST] How to solve the integral $\int \frac{4+12x}{({1-2x-3x^2})^\frac{2}{3}}\,dx$ $$\int \frac{4+12x}{({1-2x-3x^2})^\frac{1}{3}}\,dx$$
So first I tried setting $u = 1-2x-3x^2, du = -2-6x\,dx$
Which gives
$$-2\int u^\frac{-1}{3}du = \frac{-6 u^\frac{2}{3}}{2} =-3u^\frac{2}{3}$$
And ultimately:
$$-3(1-2x-3x^2)^\frac{2}{3}$$
For some odd reason when I take the derivative I'm getting something completely different, what's wrong with my substitution?
 [/INST] \begin{align}
& \frac d {dx} \left( -3(1-2x-3x^2)^{2/3} \right) \\[10pt]
= {} & -3\cdot \frac 2 3(1-2x-3x^2)^{(2/3)-1}(-2-6x) & & \text{by the chain rule} \\[10pt]
= {} & -2(1-2x-3x^2)^{-1/3}(-2-6x) \\[10pt]
= {} & (1-2x-3x^2)^{-1/3} (4+12x)
\end{align}
</s>