Type 1: integral involving sqrt(ax+b) Substitute t = sqrt(ax+b), so $t^2$ = ax+b, 2t dt = a dx.

Example: integral x*sqrt(2x+3) dx Let t = sqrt(2x+3), x = $\frac{t^2-3}{2}$, dx = t dt = integral $\frac{(t^2-3}{2}$)tt dt = $\frac{1}{2}$*integral $t^2$($t^{2-3}$) dt = $\frac{1}{2}$*integral ($t^4$ - 3$t^2$) dt = $\frac{1}{2}$($t^5$/5 - $t^3$) + C Substitute back t = sqrt(2x+3).

Type 2: integral involving sqrt$\frac{(ax+b}{(cx+d)}$) or (ax+b)^$\frac{p}{q}$ Substitute t = $\frac{(ax+b}{(cx+d)}$)^$\frac{1}{n}$ where n is the LCM of all fractional powers.

Type 3: Euler Substitutions for sqrt($ax^{2+bx+c}$)

• If a > 0: let sqrt($ax^{2+bx+c}$) = t +/- x*sqrt(a)
• If c > 0: let sqrt($ax^{2+bx+c}$) = xt +/- sqrt(c)
• If roots alpha, beta exist: let sqrt(a(x-alpha)(x-beta)) = t(x-alpha)

These are powerful but algebraically heavy — use standard forms when possible.

Type 4: integral $\frac{dx}{x^n*sqrt(ax^2+bx+c}$) Substitute x = 1/t to reduce the power of x.

Example: integral $\frac{dx}{x^2*sqrt(1+x^2}$) Let x = 1/t, dx = -dt/$t^2$. sqrt(1+1/$t^2$) = sqrt($t^{2+1}$)/|t| = integral $\frac{-dt/t^2}{(1/t^2 * sqrt(t^2+1)}$/t) = -integral t/sqrt($t^{2+1}$) dt = -sqrt($t^{2+1}$) + C = -sqrt(1+1/$x^2$)*|x|/|x| + C = -sqrt$\frac{x^2+1}{x}$ + C (for x > 0)