Which familiar derivative pairs become basic integrals?
You will be able to: Use exponential, logarithmic, trigonometric and inverse-trigonometric antiderivative pairs.
Which familiar derivative pairs become basic integrals?
A calculator’s undo operation must match the original operation. Integration also asks which function differentiates to exactly the given expression, including its sign and scale.
A useful starting point: How do we reverse differentiation without losing the constant? →
Words and symbols before equations
- ln x
- Natural logarithm, inverse of eˣ for x>0.
- abs(x)
- Absolute value, needed in ln(abs(x)) for negative x.
- arctan x
- Inverse tangent with derivative 1/(1+x²).
- arcsin x
- Inverse sine with derivative 1/√(1−x²) for −1<x<1.
What this picture assumes
Original model; readouts are rounded. Each chosen F differentiates to f. Radian angles. The logarithm panel restricts x>0; ln(abs(x)) works separately on x<0. Curves use different vertical scales.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- f=eˣ; F=eˣ. At x=0.5, f=1.64872; F=1.64872; F′=1.64872. The graphs have separate vertical scales.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
The exponential eˣ differentiates to itself, so ∫eˣdx=eˣ+C. For a>0, a≠1, ∫aˣdx=aˣ/ln(a)+C, because its derivative otherwise contributes ln(a). Also ∫(1/x)dx=ln(abs(x))+C on intervals excluding zero.
Reverse the familiar radian-based pairs: ∫cos x dx=sin x+C, ∫sin x dx=−cos x+C, ∫sec²x dx=tan x+C, and ∫sec x tan x dx=sec x+C. Work on intervals without singularities.
Likewise ∫csc²x dx=−cot x+C and ∫csc x cot x dx=−csc x+C. The negative signs are essential. Inverse-trig forms include ∫1/(1+x²)dx=arctan x+C and ∫1/√(1−x²)dx=arcsin x+C.
These are patterns, not permission to ignore an inner function’s derivative. A scaled input such as cos(3x) requires a compensating factor, developed with substitution.
A worked example, step by step
Find ∫[eˣ+2/x−sin x+1/(1+x²)]dx on x>0.
- Integrate eˣ to eˣ.
- On x>0, integrate 2/x to 2ln x.
- Since (cos x)′=−sin x, that term integrates to cos x.
- The last term gives arctan x; combine to eˣ+2ln x+cos x+arctan x+C and differentiate to check.
The integral of 1/x is logarithmic; the integral of 1/(1+x²) is inverse tangent. Similar-looking fractions need different derivative matches.
Can ln(x) replace ln(abs(x)) on a negative interval?
Compare with an explanation
No. The real logarithm ln(x) is undefined for x<0; ln(abs(x)) has derivative 1/x there.
Predict. Change one thing. Explain.
Switch among the derivative pairs in the model. At the selected input compare the integrand height with the displayed antiderivative slope. State each formula’s domain restriction.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
f=eˣ; F=eˣ. At x=0.5, f=1.64872; F=1.64872; F′=1.64872. The graphs have separate vertical scales.
Original model; readouts are rounded. Each chosen F differentiates to f. Radian angles. The logarithm panel restricts x>0; ln(abs(x)) works separately on x<0. Curves use different vertical scales.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the rate units, signed contributions, theorem conditions, domain or antiderivative check. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionIntegrate 3eˣ+sec²x−2/x on x>0. Then explain how the logarithmic term must be written on x<0.
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Compare with the answer and four-point rubric
- 1 point: 3eˣ integrates to 3eˣ.
- 1 point: sec²x integrates to tan x on an interval where it exists.
- 1 point: On x>0 the family is 3eˣ+tan x−2ln x+C.
- 1 point: On x<0 use −2ln(abs(x)); exclude zero and tangent singularities and allow a separate constant on each interval.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why is ∫sin x negative cosine?
Differentiating −cos x gives +sin x.
RECALL 2Which pair handles 1/(1+x²)?
The derivative of arctan x.
RECALL 3What restriction accompanies ln(abs(x))?
x≠0, working on an interval on one side of zero.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Which familiar derivative pairs become basic integrals?
- ∫eˣdx=eˣ+C; ∫1/x dx=ln(abs(x))+C.
- ∫cos x dx=sin x+C; ∫sin x dx=−cos x+C.
- ∫sec²x dx=tan x+C; ∫sec x tan x dx=sec x+C.
- ∫csc²x dx=−cot x+C; ∫csc x cot x dx=−csc x+C.
- ∫1/(1+x²)dx=arctan x+C; ∫1/√(1−x²)dx=arcsin x+C.
Remember: The integral of 1/x is logarithmic; the integral of 1/(1+x²) is inverse tangent. Similar-looking fractions need different derivative matches.
Conditions: Original model; readouts are rounded. Each chosen F differentiates to f. Radian angles. The logarithm panel restricts x>0; ln(abs(x)) works separately on x<0. Curves use different vertical scales.
Refresh Kid · AP Calculus BC Unit 6 · Objectives FUN-6.C · Review edition
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Mapped to College Board CED, Topic 6.8, FUN-6.C. CED effective Fall 2020 and Fall 2026 clarifications checked September 17, 2026. BC scope includes all fourteen topics, including integration by parts (6.11), nonrepeating linear partial fractions (6.12) and improper integrals (6.13). Topic 6.14 integrates the BC toolkit and Skill 1.C. Focused lesson titles, examples and questions are original Refresh Kid material.
Distinguish signed accumulation, unsigned area and initial amount. Error direction needs interval-wide shape information. FTC hypotheses, variable-bound chain factors, integration constants, transformed bounds and domain restrictions are explicit. Bounded jumps are treated separately from differentiability of accumulation. The BC extension teaches parts from the product rule, decomposition with distinct linear factors and independent improper limits. Finite truncations do not certify convergence. Shared foundation lessons are maintained with AB; BC extensions and method selection are authored for this course.
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