Integration by Partial Fraction Decomposition (with Repeated Linear Factors)

Introduction

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 decomposed into simpler fractions, which can be integrated using basic rules to be explains by the illustrative examples below.

Illustrative Example #1

Evaluate the integral \displaystyle \int \dfrac{3x^2 + 7x +1}{x^3+2x^2 +x}\,dx

First, factor the denominator:

Since the linear factor (x+1) is repeated, decompose using:

Multiply both sides by the denominator x(x+1)^2 to eliminate fractions:

Now, solve for A, B, \text{ and } C by plugging in values:

Let x=0:

Let x=-1:

To solve for B, we can plug in A=1, \, C =3, and choose a random value for x, for simplicity, we will choose x=1:

So we now have the decomposition:

Now substitute into the integral:

Integrate each term (solved using u-substitution and power rule):

Therefore,

Illustrative Example #2

Evaluate the integral \displaystyle \int \frac{5x - 2}{x^2 + 6x + 9} \, dx.

We begin by factoring the denominator

Since the linear factor (x + 3) is repeated (multiplicity 2), decompose the integrand as

Multiply through by the denominator (x + 3)^2to clear fractions:

Let x=-3

Plug-in B = -17and assign a random value of xby substituting x = 0 to obtain the value of A:

Hence the partial-fraction decomposition is

Substitute this into the integral:

Integrate each term:

Therefore,

Illustrative Example #3

Evaluate the integral \displaystyle \int \frac{14x^2 + 51x + 81}{x^3 - x^2 - 16x - 20} \, dx

First, factor the denominator:

Because the linear factor (x + 2) is repeated (multiplicity 2), decompose the fraction as

Multiply through by the common denominator (x - 5)(x + 2)^2 to clear fractions:

Let x=5:

Let x=-2:

To find B, substitute A = 14 and C = -5, and plug any convenient value; choose x = 0:

Hence the decomposition is

Substitute back into the integral:

Integrate each term:

Therefore,