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Calculus II (Integration & Series) Unit 4: Partial fractions & improper integrals

Partial fractions split rational functions into simpler pieces you can integrate. Improper integrals handle infinite limits or infinite integrands.

Covers: Partial fraction decomposition, improper integrals, comparison for integrals

Study guide

Partial fraction decomposition breaks a rational function into simpler fractions that integrate easily. First check degrees: if the numerator's degree is at least the denominator's, divide first. Then factor the denominator. Each distinct linear factor (x − a) gets a term A/(x − a); a repeated factor (x − a)² gets A/(x − a) + B/(x − a)²; an irreducible quadratic factor gets (Ax + B)/(x² + bx + c). Solve for the constants by clearing denominators and plugging in convenient x-values or matching coefficients. Linear terms integrate to logarithms, and quadratic ones often to arctangents.

An improper integral has an infinite limit or an integrand that blows up inside the interval. Define it as a limit: ∫ₐ^∞ f(x) dx = lim b→∞ ∫ₐᵇ f(x) dx. If the limit is finite, the integral converges; otherwise it diverges. For a vertical asymptote at an endpoint or inside the interval, approach it with a limit from the correct side and split at the asymptote if needed.

The p-integrals are benchmarks: ∫₁^∞ 1/xᵖ dx converges if p > 1 and diverges if p ≤ 1. So ∫₁^∞ 1/x² dx = 1, but ∫₁^∞ 1/x dx diverges.

The comparison test decides convergence without evaluating: if 0 ≤ f ≤ g and ∫ g converges, so does ∫ f; if ∫ f diverges, so does ∫ g. Compare to p-integrals or exponentials.

For a distinct linear factor (x − a), find its constant A by covering (x − a) in the original fraction and evaluating what remains at x = a. For a repeated factor, this method gives the constant on its highest power. To integrate (Ax + B)/(x² + c²), split it into A·x/(x² + c²), which integrates to (A/2)ln(x² + c²), plus B/(x² + c²), which integrates to (B/c)arctan(x/c). Complete the square first if the quadratic has an x-term.

Near zero the p-integral rule reverses: ∫₀¹ 1/xᵖ dx converges when p < 1 and diverges when p ≥ 1. Evaluating improper integrals often needs limits such as lim b→∞ b e^(−b) = 0 (by L'Hospital's rule) and lim b→∞ arctan b = π/2.

If f and g are positive and lim x→∞ f(x)/g(x) is a finite positive number, then ∫ₐ^∞ f and ∫ₐ^∞ g either both converge or both diverge.

Key ideas

• If the top's degree ≥ the bottom's, divide first.
• Each linear factor (x − a) gets A/(x − a); repeated factors get one term per power.
• Irreducible quadratic factors get (Ax + B)/(x² + …).
• Improper integrals are limits: replace ∞ (or the bad point) with a variable and take the limit.
• ∫₁^∞ 1/xᵖ dx converges only if p > 1.

Formulas and rules

Common mistakes

• Skipping long division when the fraction is improper.
• Integrating straight across a vertical asymptote inside the interval.

Words to know

Partial fractions (distinct linear factors)

1/((x − a)(x − b)) = A/(x − a) + B/(x − b)

Improper rational function

Numerator degree ≥ denominator degree; divide before partial fractions.

Convergent improper integral

∫₁^∞ 1/xᵖ dx converges when p > 1 (p = 2 gives 1).

Divergent improper integral

∫₁^∞ 1/x dx grows without bound.

Improper integral (infinite limit)

∫ₐ^∞ f(x) dx = lim b→∞ ∫ₐᵇ f(x) dx

Partial fraction decomposition

Writing a rational function as a sum of simpler fractions.

Repeated linear factor

(x − a)² gets A/(x − a) + B/(x − a)².

Irreducible quadratic factor

Gets (Ax + B)/(x² + bx + c).

Improper integral

An integral with an infinite limit or an infinite discontinuity.

Converges / diverges

An improper integral converges if its limit is finite.

Comparison test (integrals)

If 0 ≤ f ≤ g and ∫g converges, so does ∫f.

p-integral

∫₁^∞ 1/xᵖ dx converges if p > 1.

Sample practice questions

29 more questions like these, from level 1 to 5, are in the practice set.