Integration by Partial Fraction Decomposition (with Irreducible Quadratic Factors)
June 28, 2025
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The method of partial fractions is a powerful technique used to integrate rational functions in proper form (that is, the degree of the numerator is less than the degree of the denominator), which are expressions of the form \dfrac{f(x)}{g(x)} where f(x) \text{ and } g(x) are polynomials. Once in proper form, the rational expression is then decompose into simpler fractions, which can be integrated using basic rules to be explains by the illustrative examples below.
If x^2+px+q is an irreducible quadratic factor of the denominator g(x), and its highest power dividing g(x) \text{ is } (x^2+px+q)^n, then assign a partial fraction for each power from 1 \text{ to } n. Each term should have a linear numerator of the form Bx+C. Specifically, write:
Repeat this process for each distinct irreducible quadratic factor found in g(x).
This method is applicable when the degree of the numerator is less than the degree of the denominator. If it is not, polynomial long division should be performed first to rewrite the function in proper form.
Illustrative Example #1
Evaluate the integral \displaystyle \int \dfrac{10}{x^3-9x^2+9x-1} \, dx
We begin by factoring the denominator:
We then multiply both sides by (x-1)(x^2+9) to eliminate denominators:
Choose convenient values of x to solve for A, B, \text{and } C. Let’s begin with substituting x = 1:
This simplifies the equation to:
We can now solve algebraically to obtain the remaining values:
Thus, we have
We then integrate each term:
Note that we can solve each integral using u-sub, and some common integrals:
Thus,
Illustrative Example #2
Evaluate the integral \displaystyle \int\dfrac{x^2-x+6}{x^3+3x} \, dx
We need to factor the denominator:
With the denominator being factored out, we can now set up the partial fractions of the integrand:
We then multiply both sides by (x)(x^2 + 3) to eliminate denominators:
Choose convenient values of x to solve for A, B, \text{ and }C. Let's begin with substituting x = 0:
This simplifies the equation to:
We can now solve algebraically to obtain the remaining values:
Hence, we have the partial fraction complete:
We then integrate each term: