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CalculusBy Educator Today

C27 · Unit C6 Integration and accumulation

Linear partial fractions

BC extensionReview editionNot yet human-reviewed
Time
About 45 minutes, untimed
Course framework reference
AP 6.12, 6.14 · BC
Builds on
Part of

Progress on this lesson counts in every pathway listed.

Understand the idea

Partial fractions decomposes a proper rational function into simpler terms. For distinct linear factors, assign one constant numerator to each factor, multiply by the common denominator and equate coefficients or convenient values. Perform polynomial division first if the numerator degree is not smaller. Each reciprocal linear term integrates to a logarithm with its chain factor.

Worked example

1/[(x+1)(x+2)] equals A/(x+1)+B/(x+2). Multiplication gives 1 equals A(x+2)+B(x+1). Set x equals minus 1 and minus 2 to obtain A equals 1,B equals minus 1. The integral is ln|x+1| minus ln|x+2|+C on intervals avoiding both poles.

Explore the mathematics

Linear partial fractions

Decompose a rational integrand into simpler fractions.

Setting 3 of 5

Scroll sideways to see the full diagram.

Can a rational function be split into simpler fractions?xy1Marked input 0.5; displayed curve height 0.2667∫₀¹ 1/[(x+1)(x+2)] dx

Predict before continuing

Can a rational function be split into simpler fractions?

Read the complete explanation
  1. Can a rational function be split into simpler fractions?
  2. 1/[(x+1)(x+2)] = 1/(x+1)−1/(x+2).
  3. Integrate to ln|x+1|−ln|x+2|.
  4. On [0,1] the result is ln(4/3). Poles at −1 and −2 are outside this interval.
Linear partial fractions · visual worked method
Linear partial fractions · visual worked method1/[(x+1)(x+2)]=A/(x+1)+B/(x+2). Multiplication gives 1=A(x+2)+B(x+1). Set x=−1 and −2 to obtain A=1,B=−1. The integral is ln|x+1|−ln|x+2|+C on intervals avoiding both poles.Worked example · 1/21/[(x+1)(x+2)]=A/(x+1)+B/(x+2).Multiplication gives1=A(x+2)+B(x+1). Set x=−1 and −2to obtain A=1,B=−1. The integralisWorked example · 2/2ln|x+1|−ln|x+2|+C on intervalsavoiding both poles.

1/[(x+1)(x+2)]=A/(x+1)+B/(x+2). Multiplication gives 1=A(x+2)+B(x+1). Set x=−1 and −2 to obtain A=1,B=−1. The integral is ln|x+1|−ln|x+2|+C on intervals avoiding both poles.

Discuss: explain what each label represents, then draw a related example with different values.

A common misconception

Partial fractions does not remove domain exclusions. Logarithms of absolute values are needed when intervals permit negative factors.

Investigate and explain

Check a decomposition by recombining it before integrating. Explain why this is a stronger check than evaluating at one input.
Identify one step in the worked example, name the mathematical fact that permits it, and state its domain conditions.