Calculus Of One Real Variable By Pheng Kim Ving
Chapter 10: Techniques Of Integration Section 10.7: The Method Of Partial Fractions

 

10.7
The Method Of Partial Fractions

 

 

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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. 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 Ai'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:

 

 

Solving the system of equations:

 

 

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
Ai's, i = 1, 2, ..., m, Bj's, and Cj'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).

 

ii. Factor the denominator Q(x) into linear and/or irreducible quadratic factors.

 

iii. Perform the partial-fraction expansion on P(x)/Q(x), or on R(x)/Q(x) if part i is carried out.

 

iv. Integrate the resulting expression of P(x)/Q(x).

 

Note On Long Division

 

For example, given:

 

 

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Problems & Solutions

 

1. Calculate the following integrals.

 

 

Solution

 

 

 

 

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2. Compute the following integrals.

 

 

Solution

 

 

 

 

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3. Evaluate:

 

 

Solution

 

 

 

where C = (1/2)C1.

 

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4. Find:

 

 

Solution

 

 

 

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5. Calculate:

 

 

Solution

 

Let u = ex. Then du = ex dx = u dx, yielding dx = (du)/u. So:

 

 

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