Calculus I · Unit 3A · lesson
Partial Fractions
Learning objectives
Decompose proper rational functions and integrate linear and repeated-linear terms.
Partial Fractions
Turn one rational function into several familiar ones
Partial fractions is an algebraic decomposition used before integration. A proper rational function with a factorable denominator can often be written as a sum of simpler fractions whose antiderivatives are logarithmic or arctangent forms. The calculus step is usually easy once the decomposition is correct; most errors originate in incomplete factoring or an incorrect template.
Begin by performing polynomial division if the numerator's degree is not smaller than the denominator's. Then factor over the real numbers and include the correct terms for repeated linear factors and irreducible quadratics. Solve for the coefficients, integrate term by term, and differentiate or recombine the result as a check. Treating the algebra as part of the method, rather than as preliminary clutter, makes the procedure far more reliable.
When a rational function is proper and the denominator factors, rewrite it as a sum of simpler fractions.
Two distinct linear factors
Evaluate
Worked solution
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Before decomposing
If the numerator degree is at least the denominator degree, perform polynomial division first. Then factor the denominator completely over the real numbers and include the correct form for repeated or irreducible quadratic factors.
u3a-pf-01Evaluate .
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Decompose as .
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After the explanation
Use the section idea
Choose an integration method from the integrand's structure, then verify the result by differentiation.
Substitution reverses a chain rule, parts reverses a product rule, and algebraic or trigonometric rewrites expose a recognizable derivative pattern.
Simplify first; look for an inner derivative; then consider parts, identities, trigonometric substitution, or partial fractions in a deliberate order.
Choosing a method by surface appearance, transforming only part of the differential, or accepting a more complicated integral than the one you started with.
Can you name the derivative rule being reversed and differentiate the final answer back to the integrand?
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