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 = (t^2-3)/2, dx = t dt = integral ((t^2-3)/2)tt dt = (1/2)*integral t^2(t^2-3) dt = (1/2)*integral (t^4 - 3t^2) dt = (1/2)(t^5/5 - t^3) + C Substitute back t = sqrt(2x+3).

Type 2: integral involving sqrt((ax+b)/(cx+d)) or (ax+b)^(p/q) Substitute t = ((ax+b)/(cx+d))^(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 dx/(x^n*sqrt(ax^2+bx+c)) Substitute x = 1/t to reduce the power of x.

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