$$\begin{gathered}\int x^n = \frac{x^{n+1}}{n+1}+C\\[3pt] \int(ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)}+C\end{gathered}$$

$6x^2+1$ and $6x^2+5$ both differentiate to $12x$ — integration can't tell which constant was there, so always add $+C$.

Reflex: Power add 1, ÷ new power, ÷ inner coefficient, +C. Every time.

$$\begin{gathered}x(3x^2-1)^2 \to 9x^5-6x^3+x\\[2pt] \tfrac{6x^8+7x^2+5x}{2x^5} \to 3x^3 + \tfrac{7}{2}x^{-3} + \tfrac{5}{2}x^{-4}\\[2pt] \sqrt{7-3x} = (7-3x)^{1/2}\\[2pt] \left(\tfrac{2}{3-7x}\right)^3 = 8(3-7x)^{-3}\end{gathered}$$

Integration has no product, quotient, or chain rule — expand, divide out, or rewrite as powers first.

Reflex: Make it a sum of $ax^n$/$(ax+b)^n$ terms; the only 'technique' is algebra first.

$$\begin{gathered}\int\sin(ax+b) = \tfrac{-\cos(ax+b)}{a}\\[2pt] \int\cos = \tfrac{+\sin}{a}, \quad \int\sec^2 = \tfrac{\tan}{a}\\[3pt] \sin^2x = \tfrac{1-\cos 2x}{2}, \quad \cos^2x = \tfrac{1+\cos 2x}{2}\\[2pt] \tan^2x = \sec^2x - 1\end{gathered}$$

Linear angles integrate directly; squares don't. Reflex: See a squared trig under ∫ → identity first; ∫tan doesn't exist here.

$$\int e^{ax+b}dx = \frac{e^{ax+b}}{a}+C$$ $$\int\frac{k}{ax+b}dx = \frac{k\ln(ax+b)}{a}+c$$

Rewrite first: $\frac{4}{e^{7-6x}} = 4e^{6x-7}$, $\sqrt{e^{-x}} = e^{-x/2}$, $e^x\sqrt{e^{3x}} = e^{5x/2}$. No AM integral for $\ln x$.

Reflex: Exponentials: copy, ÷ inner coefficient. ln: differentiate yes, integrate no.

$$\int(5x+1)^6dx = \frac{(5x+1)^7}{7(5)}+c$$ $$\int\frac{dx}{2x+3} = \frac{\ln(2x+3)}{2}+c$$

Power +1, ÷ new power ÷ inner coefficient, +c. Power −1 breaks this (÷0) → ln instead. Reflex: Up-one-divide, ÷ inner coefficient, +c; power −1 → ln.