Type 1: Infinite Limits integral(a to infinity) f(x) dx = lim(b->infinity) integral(a to b) f(x) dx

Type 2: Integrand has vertical asymptote at c in [a,b] integral(a to b) f(x) dx = integral(a to c-) f(x) dx + integral(c+ to b) f(x) dx (both as limits)

JEE Note: Full improper integral theory is beyond JEE Main scope, but knowing that integral(0 to 1) 1/sqrt(x) dx = 2 (converges) and integral(0 to 1) 1/x dx diverges is useful.

Trap: Do NOT blindly apply FTC if the integrand is discontinuous within [a,b]. For example, integral(-1 to 1) 1/$x^2$ dx is NOT [-1/x] from -1 to 1 = -2. The integral diverges because 1/$x^2$ has a non-integrable singularity at x = 0.