Antiderivatives and Indefinite Integrals

Find antiderivatives of common functions for AP Calculus AB.

An antiderivative of f(x)f(x) is a function F(x)F(x) such that F′(x)=f(x)F'(x) = f(x). The indefinite integral represents the family of all antiderivatives.

Basic Antiderivatives

∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C (n≠−1n \neq -1)

∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln|x| + C

∫ex dx=ex+C\int e^x\,dx = e^x + C, ∫ax dx=axln⁡a+C\int a^x\,dx = \frac{a^x}{\ln a} + C

∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C, ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C

∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C, ∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x\,dx = -\cot x + C

∫11+x2 dx=arctan⁡x+C\int \frac{1}{1+x^2}\,dx = \arctan x + C, ∫11−x2 dx=arcsin⁡x+C\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C

Initial Value Problems

Given f′(x)f'(x) and f(a)=bf(a) = b: integrate, then use the condition to find CC.

Worked Example

f′(x)=3x2−2f'(x) = 3x^2 - 2, f(1)=4f(1) = 4.

f(x)=x3−2x+Cf(x) = x^3 - 2x + C. f(1)=1−2+C=4f(1) = 1 - 2 + C = 4 → C=5C = 5.

f(x)=x3−2x+5f(x) = x^3 - 2x + 5.

Practice Problems

    1. ∫(4x3−2x+7) dx\int (4x^3 - 2x + 7)\,dx.
    1. ∫(sin⁡x+ex) dx\int (\sin x + e^x)\,dx.
    1. f′(x)=cos⁡xf'(x) = \cos x, f(0)=3f(0) = 3. Find f(x)f(x).

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Key Takeaways

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    Antiderivative = reverse of derivative.

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    Always include +C+ C for indefinite integrals.

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    Use initial conditions to find CC.

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