How can two simpler fractions unlock a rational integral?
You will be able to: Decompose a proper rational function with distinct linear factors and integrate the pieces.
How can two simpler fractions unlock a rational integral?
Adding 2/3 and 1/4 first requires a common denominator. Partial fractions reverses that familiar process: start with one fraction and recover the simpler pieces.
A useful starting point: How do bounds and repeated steps work in integration by parts? →
Words and symbols before equations
- Rational function
- A quotient of polynomials.
- Proper fraction
- Numerator degree is less than denominator degree.
- Distinct linear factors
- Different first-degree factors, such as x+1 and x+2.
- Identity
- An equality true at every allowed input, not just a few samples.
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
For (3x+5)/((x+1)(x+2)), write A/(x+1)+B/(x+2). Multiplying by the common denominator gives 3x+5=A(x+2)+B(x+1).
Comparing coefficients gives A+B=3 and 2A+B=5, so A=2 and B=1. Substituting x=−1 or −2 into the polynomial identity is also valid for finding coefficients; the rational expression itself is still undefined there.
Integrate the pieces: 2ln(abs(x+1))+ln(abs(x+2))+C. The derivative recovers the original fraction on each interval excluding −2 and −1.
First divide if the fraction is not proper. This AP topic focuses on nonrepeating linear factors; repeated factors and irreducible-quadratic decompositions are outside this lesson’s assessed scope.
A worked example, step by step
Find ∫1/((x+1)(x+3))dx.
- Set 1=A(x+3)+B(x+1).
- The x coefficient gives A+B=0 and constants give 3A+B=1, hence A=1/2, B=−1/2.
- Integrate: (1/2)ln(abs(x+1))−(1/2)ln(abs(x+3))+C.
- Differentiate and combine: ((x+3)−(x+1))/(2(x+1)(x+3)) equals the original integrand; x≠−1,−3.
The numerators A and B are unknown constants to solve for. Splitting the numerator across factors without a valid identity changes the function.
Does using x=−1 in the coefficient identity put −1 back in the rational domain?
Compare with an explanation
No. It solves a polynomial identity; the original denominator exclusion remains.
Predict. Change one thing. Explain.
Move x through [0,4]. Compare the original rational value and the sum 2/(x+1)+1/(x+2). Use the identity, rather than sampled agreement alone, to justify equality.
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 questionDecompose and integrate (3x+4)/((x+1)(x+2)), stating domain restrictions.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Set 3x+4=A(x+2)+B(x+1).
- 1 point: A+B=3 and 2A+B=4 give A=1, B=2.
- 1 point: An antiderivative is ln(abs(x+1))+2ln(abs(x+2))+C.
- 1 point: The original function excludes −1 and −2; differentiating verifies the result separately on each allowed 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 1How are the constants found?
Clear denominators and solve the resulting polynomial identity.
RECALL 2What must be preserved?
Every original denominator exclusion.
RECALL 3Which factors are emphasized here?
Distinct, nonrepeating linear factors.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can two simpler fractions unlock a rational integral?
- P(x)/((x−a)(x−b))=A/(x−a)+B/(x−b), for distinct a,b and proper degree.
- ∫dx/(x−a)=ln(abs(x−a))+C on intervals avoiding a.
Remember: The numerators A and B are unknown constants to solve for. Splitting the numerator across factors without a valid identity changes the function.
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.
Want to work through this with a tutor?
Bring your question about How can two simpler fractions unlock a rational integral? Your explanation and answers remain free to access.