CourseCourseQ BankQuestion BankDocsDocuments

6.12 Integrating Using Linear Partial Fractions

Syllabus
2020
Topic
6.12
Level
—

Learning objectives

—FUN-6.F—For integrands requiring integration by linear partial fractions: (a) Determine indefinite integrals. (b) Evaluate definite…• FUN-6.F For integrands requiring integration by linear partial fractions: (a) Determine indefinite integrals. (b) Evaluate definite integrals.• FUN-6.F.1 Some rational functions can be decomposed into sums of ratios of linear, nonrepeating factors to which basic integration techniques can be applied.• Enduring understanding FUN-6: Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.

Split Rational Functions into Integrable Fractions

Use linear partial fractions when a rational function is proper (numerator degree is less than denominator degree) and the denominator factors into distinct linear factors. For (x−a)(x−b)(x-a)(x-b) with a≠ba\ne b, write P(x)(x−a)(x−b)=Ax−a+Bx−b\frac{P(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}, then solve for AA and BB.

Step Action Purpose
1 if necessary, use long division make the rational part proper
2 factor the denominator identify each distinct linear factor
3 assign one constant numerator per factor build the partial-fraction form
4 multiply through by the denominator and solve determine the constants
5 integrate each simple fraction obtain logarithmic terms

Decompose 3x+5(x+1)(x+2)=Ax+1+Bx+2\frac{3x+5}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}. Multiplying through gives 3x+5=A(x+2)+B(x+1)3x+5=A(x+2)+B(x+1), so A=2A=2 and B=1B=1. Therefore ∫3x+5(x+1)(x+2) dx=2ln⁡∣x+1∣+ln⁡∣x+2∣+C\int\frac{3x+5}{(x+1)(x+2)}\,dx=2\ln|x+1|+\ln|x+2|+C.

Since 1(x+1)(x+2)=1x+1−1x+2\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}, ∫01dx(x+1)(x+2)=[ln⁡(x+1)−ln⁡(x+2)]01=ln⁡(4/3)\int_0^1\frac{dx}{(x+1)(x+2)}=[\ln(x+1)-\ln(x+2)]_0^1=\ln(4/3). The interval contains no denominator zero, so the integrand is continuous there.

This card covers distinct, nonrepeating linear factors only. Do not use the same template unchanged for repeated factors or irreducible quadratic factors. Keep ln⁡∣x−a∣\ln|x-a| in an indefinite result, and remember that decomposition does not remove excluded values from the original denominator.