How do logarithmic pieces combine in a definite integral?
You will be able to: Evaluate definite partial-fraction integrals on intervals without poles.
How do logarithmic pieces combine in a definite integral?
A total built from two rates can be found by adding their individual totals. A rational rate written as a sum of simpler fractions works the same way, provided neither piece becomes unbounded inside the interval.
A useful starting point: How can two simpler fractions unlock a rational integral? →
Words and symbols before equations
- Pole
- An input where the rational function becomes unbounded.
- Logarithm difference
- ln(b)−ln(a)=ln(b/a) for positive a,b.
- Endpoint evaluation
- Apply F(b)−F(a) to the full antiderivative.
- Interval check
- Verify the integrand remains defined and continuous throughout the interval.
What this picture assumes
Original BC model; rounded readouts. f=(3x+5)/((x+1)(x+2))=2/(x+1)+1/(x+2). Display domain [0,4] avoids poles −1 and −2. Integral from 0 to b subtracts F(0)=ln2.
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.
- At b=1, original=1.33333 and component sum=1.33333. Integral 0→b=1.79176. Equality comes from the polynomial identity, not just plotted samples.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
On [0,1], (3x+5)/((x+1)(x+2)) is continuous. Its decomposition 2/(x+1)+1/(x+2) can be integrated term by term.
Use F=2ln(x+1)+ln(x+2) on this interval. F(1)=2ln2+ln3 and F(0)=ln2. The difference is ln6.
Absolute values in a general antiderivative protect logarithm domains, but do not permit ordinary FTC evaluation across a pole. For example [−3,0] crosses both excluded inputs of this rational expression.
When a bound interval crosses an unbounded point, split it and use independent one-sided limits. The improper-integral lessons explain why cancellation cannot repair divergence.
A worked example, step by step
Evaluate ∫₀¹1/((x+1)(x+3))dx.
- There are no poles on [0,1]; the integrand is continuous and positive.
- Use F=(1/2)ln(x+1)−(1/2)ln(x+3).
- F(1)−F(0)=(1/2)(ln2−ln4+ln3).
- Combine to (1/2)ln(3/2)≈0.203, which is positive as required.
Subtract the entire lower-endpoint expression. A finite logarithmic formula at the endpoints does not prove the interval between them is safe.
Can F(0) be ignored because the lower bound is zero?
Compare with an explanation
No. Logarithms evaluated at x=0 may be nonzero; here F(0)=ln2.
Predict. Change one thing. Explain.
Move b from 0 to4. Read the exact accumulated value from 0 to b. Compare its increasing behavior with the positive rational rate.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At b=1, original=1.33333 and component sum=1.33333. Integral 0→b=1.79176. Equality comes from the polynomial identity, not just plotted samples.
Original BC model; rounded readouts. f=(3x+5)/((x+1)(x+2))=2/(x+1)+1/(x+2). Display domain [0,4] avoids poles −1 and −2. Integral from 0 to b subtracts F(0)=ln2.
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 questionEvaluate ∫₀¹(3x+4)/((x+1)(x+2))dx after decomposing it.
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Compare with the answer and four-point rubric
- 1 point: The decomposition is 1/(x+1)+2/(x+2).
- 1 point: There are no poles on [0,1], so FTC applies.
- 1 point: F(1)−F(0)=ln2+2ln3−2ln2.
- 1 point: The result is ln(9/2), positive; differentiating F verifies the integrand.
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 check the interval first?
An internal singularity changes the integral into an improper one.
RECALL 2When can logarithms be combined?
When their arguments obey the logarithm identities and domains.
RECALL 3What is the lower endpoint contribution here?
F(0), which need not equal zero.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do logarithmic pieces combine in a definite integral?
- For a continuous rational integrand on [a,b], evaluate the full antiderivative as F(b)−F(a).
- If an unbounded point lies inside, ordinary endpoint subtraction is insufficient.
Remember: Subtract the entire lower-endpoint expression. A finite logarithmic formula at the endpoints does not prove the interval between them is safe.
Conditions: Original BC model; rounded readouts. f=(3x+5)/((x+1)(x+2))=2/(x+1)+1/(x+2). Display domain [0,4] avoids poles −1 and −2. Integral from 0 to b subtracts F(0)=ln2.
Refresh Kid · AP Calculus BC Unit 6 · Objectives FUN-6.F · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.12, FUN-6.F. 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.
The Organic Chemistry Tutor Fundamental Theorem of Calculus Part 1 video title, creator and description were checked; the full video was not reviewed. Khan Academy’s unit destination was checked, but its JavaScript lesson contents were not fully readable by the research tool. OpenStax Sections 5.2, 5.3 and 5.5 were consulted for conceptual cross-checking. No provider scripts, questions, diagrams or artwork were copied. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection informed optional spatial inspection and camera controls. The original tank geometry uses existing self-hosted Three.js with its MIT license. Its 2×2 dm base and water depth use the same volume as the rate calculation: 1 dm³=1 L. The initial water is blue and newly accumulated inflow is teal. The rate graph is an abstract amount calculation; it is not the physical shape of water. Camera rotation changes only the view. Full 2D graphs, readouts and explanations remain available without WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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