6.5.2 The Method Of Partial Fractions

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Calculus 1 Problems & Solutions – Chapter 6 – Section 6.5.2

6.5.2
The Method Of Partial Fractions

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Review

1. Notes

Recall from algebra that a linear function is a polynomial of degree 1, ie a function of the form ax + b (its graph is a
line). A quadratic function is a polynomial of degree 2, ie a function of the form ax² + bx + c. A real-valued polynomial is
said to be irreducible if it can't be factored. For example, x² – x – 2 can be factored as (x + 1)(x – 2), while x² + x + 1
can't be factored, so is irreducible. Recall that ax² + bx + c is irreducible if b² – 4ac < 0. Note that all real-valued linear
functions are irreducible.

A rational function is a ratio or fraction P(x)/Q(x) where P(x) and Q(x) are polynomials.

In this section we're concerned with the integration of rational functions. A rational function may not readily lend itself to
a substitution method. If that's the case, it'll be expressed as a sum of simpler fractions, known as partial fractions, which
are easier to integrate.

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2. Partial Fractions – Linear Factors

Consider, for example, the rational function:

Indeed it's correct.

Another method of determining A and B is as follows. Multiplying both sides of Eq. [2.1] by the denominator x + 1 below
A we obtain:

Case Of n Distinct Linear Factors

In general, if the degree of the numerator P(x) is less than that of the denominator Q(x) and if Q(x) factors into a
product of n distinct linear factors, say:

The constants A_i's, i = 1, 2, ..., n, can be determined by the add-up-the-partial-fractions method or the limit-procedure
method as in the above example, where n = 2.

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3. Partial Fractions – Quadratic Factors

Consider as another example the rational function:

Here, the limit-procedure method can be used to determine A, but there's no simple way to use it to determine B or C.

Remark 3.1

You may ask why we don't use a constant numerator for a partial fraction with a quadratic denominator to make things
simpler, like this: A/(x + 2) + B/(x² + x + 1). Well, let's see:

which has no solutions. That is, There are no constants A and B such that the given rational function can be expanded to
A/(x + 2) + B/(x² + x + 1). That's the answer to the question.

Case Of m Distinct Linear Factors And n Distinct Quadratic Factors

In general, if the degree of the numerator P(x) is less than that of the denominator Q(x) and if Q(x) factors into a
product of m distinct linear factors and n distinct irreducible quadratic factors, say:

Again corresponding to a linear denominator we use a constant numerator and corresponding to a quadratic denominator
we use a linear numerator. That is, the degree of the numerator is less than that of the denominator by 1. The constants
A_i's, i = 1, 2, ..., m, B_j's, and C_j's, j = 1, 2, ..., n, can be determined by the add-up-the-partial-fractions method as in the
above example, where m = n = 1.

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4. Partial Fractions – Multiplicity

Consider, for example, the rational function:

Since the multiplicity of the factor x is 4, there are 4 partial fractions corresponding to x, with denominators having
exponents increasing from 1 to 4. There's only 1 partial fraction corresponding to x – 3, and there are 3 corresponding to
x² + 5, with denominators' exponents increasing from 1 to 3.

The constants A₁, A₂, A₃, A₄, B, C₁, C₂, C₃, D₁, D₂, and D₃ can be determined by the add-up-the-partial-fractions
method.

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5. Partial Fractions – General Case

The following theorem of polynomial algebra summarizes the general case of the partial-fraction expansion of a rational
function.

Theorem 5.1

Let Q(x) be a polynomial. Then Q(x) can be factored into a product of a constant, linear factors, and irreducible
quadratic factors, as follows:

The proof of this theorem is omitted because it appropriately belongs to the domain of polynomial algebra. Here we simply
utilize the theorem.

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6. The Method Of Partial Fractions

Example 6.1

Find:

Solution

EOS

Procedure

Suppose we are to find the integral:

If we don't know how to do it, we decompose P(x)/Q(x) into a sum of partial fractions and integrate the resulting
expression. This technique is called the method of partial fractions. Its procedure is summarized as follows:

i. If the degree of P(x) is greater than or equal to that of Q(x), use polynomial long division to divide P(x) by Q(x) to
obtain P(x)/Q(x) = q(x) + R(x)/Q(x) (from P(x) = q(x)Q(x) + R(x)), where q(x) is the quotient, R(x) is the
remainder, and the degree of R(x) is less than that of Q(x).