Integrating Partial Fractions: Split Distinct Linear Factors into Logarithms

For a proper rational expression with distinct linear denominator factors, split it into simpler fractions, solve their numerators, then integrate each term. Check the decomposition by recombining it.

Prepare the rational expression first

A proper rational expression has numerator degree smaller than denominator degree. If the degrees are equal or the numerator degree is larger, perform polynomial division first. Factor the denominator and record the excluded real inputs before decomposing the remaining proper fraction.

Distinct linear factors need one constant numerator each. Repeated factors require a term for each power, and irreducible quadratic factors require a different numerator form. The worked example here uses the simplest distinct-linear case so the coefficient calculation remains visible.

Worked example

Integrate (5x + 7)/[(x + 1)(x + 2)]. Write the integrand as A/(x + 1) + B/(x + 2), on intervals avoiding x = −1 and x = −2. Recombining gives numerator A(x + 2) + B(x + 1).

Match coefficients: A + B = 5 and 2A + B = 7. Subtracting the equations gives A = 2, then B = 3. Thus the integral is ∫[2/(x + 1) + 3/(x + 2)] dx.

An antiderivative is 2 ln|x + 1| + 3 ln|x + 2| + C. Differentiating returns 2/(x + 1) + 3/(x + 2), and recombining gives the original numerator 5x + 7. The formula applies on any interval that does not cross an excluded input.

Why absolute values appear

The derivative of ln|x − a| is 1/(x − a) wherever x ≠ a. Absolute values let one logarithmic formula describe either sign interval. They do not define the original integrand at its singular points or make an improper integral automatically finite.

For repeated factors, use all necessary powers. An expression with denominator (x + 3)² may require A/(x + 3) + B/(x + 3)². The second term integrates by a power rule, not a logarithm: ∫(x + 3)^(−2)dx = −1/(x + 3) + C.

Common mistake

Do not split a numerator by simply assigning its first term to the first denominator factor. The constants are determined by an identity valid for every permitted x. Recombining the proposed fractions is the fastest way to detect an incorrect assignment.

Keep decomposition and integration as separate steps. A correct fraction split can still be followed by a missing logarithm coefficient or a sign error. Check the final antiderivative by differentiating, and do not evaluate across singularities as though the integrand were continuous.

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Three practice questions with worked answers

1. Integrate 1/[(x + 1)(x + 2)].

   Decompose as 1/(x + 1) − 1/(x + 2). The answer is ln|x + 1| − ln|x + 2| + C on intervals excluding −1 and −2.

2. Integrate 1/(x² − 1).

   Use (1/2)[1/(x − 1) − 1/(x + 1)]. The result is (1/2)ln|x − 1| − (1/2)ln|x + 1| + C, excluding ±1.

3. Integrate 1/(x + 3)².

   Use the power rule on (x + 3)^(−2). The antiderivative is −1/(x + 3) + C on intervals excluding −3. Differentiation returns the positive reciprocal square.

Related study guides

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Partial Fractions. No textbook or exam-board affiliation is claimed.