SL 5.10—Indefinite integration

Syllabus
First assessment 2021
Objective
—
Level
HL

Indefinite integration reconstructs a family of functions

Indefinite integration reconstructs a family of functions.

Reverse the derivative term by term and retain the arbitrary constant; the constant represents the unknown vertical shift.

Example

∫(2x+sin x)dx=x²−cos x+C.

Differentiate the answer and use any initial value to fix C.

Do not omit C when no initial condition has selected one antiderivative.

Formula set: ∫xn dx=xn+1/(n+1)+C\int x^n\,dx=x^{n+1}/(n+1)+C for n≠−1n\ne-1, ∫dx/x=ln⁡∣x∣+C\int dx/x=\ln|x|+C, ∫ex dx=ex+C\int e^x\,dx=e^x+C, ∫sin⁡x dx=−cos⁡x+C\int\sin x\,dx=-\cos x+C, and ∫cos⁡x dx=sin⁡x+C\int\cos x\,dx=\sin x+C. For a linear composite, account for the inner gradient: ∫cos⁡(2x+3) dx=12sin⁡(2x+3)+C\int\cos(2x+3)\,dx=\tfrac12\sin(2x+3)+C. Reverse-chain recognition gives ∫kg′(x)[f(g(x))] dx\int k g'(x)[f(g(x))]\,dx by substitution or inspection.