Type 1 (a > 0): Set sqrt($ax^{2+bx+c}$) = t - sqrt(a)*x (or t + sqrt(a)*x). Squaring: $ax^{2+bx+c}$ = $t^2$ - 2sqrt(a)*tx + $ax^2$. Cancel $ax^2$: bx+c = $t^{2-2sqrt}$(a)*tx. Solve for x: x = $\frac{t^2-c}{(b+2sqrt(a)}$*t). This rationalizes the integral.

Type 2 (c > 0): Set sqrt($ax^{2+bx+c}$) = tx + sqrt(c) (or tx - sqrt(c)). Similar algebra gives a rational expression in t.

Type 3 (real roots): If $ax^{2+bx+c}$ = a(x-alpha)(x-beta), set sqrt(a(x-alpha)(x-beta)) = t(x-alpha). Then sqrt(a)(x-beta) = t^2$$\frac{x-alpha}{sqrt}(a(x-beta))... This works when the quadratic has real roots.

When to use: These are last-resort substitutions for integrals involving sqrt(quadratic) that don't yield to trigonometric substitution easily. Rare in JEE Main but useful for JEE Advanced.