7

Techniques of Integration

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7.4

Integration of Rational Functions by Partial Fractions

Copyright © Cengage Learning. All rights reserved.

Integration of Rational Functions by Partial Fractions (1 of 2)

In this section we show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate.

To illustrate the method, observe that by taking the fractions

to a common denominator we obtain

3

Integration of Rational Functions by Partial Fractions (2 of 2)

If we now reverse the procedure, we see how to integrate the function on the right side of this equation:

4

The Method of Partial Fractions

5

The Method of Partial Fractions (1 of 14)

To see how the method of partial fractions works in general, let’s consider a rational function

where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper.

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The Method of Partial Fractions (2 of 14)

We know that if

where a_n ≠ 0, then the degree of P is n and we write deg(P) = n.

If f is improper, that is, deg(P) ≥ deg(Q), then we must take the preliminary step of dividing Q into P (by long division) until a remainder R (x) is obtained such that deg(R) < deg(Q).

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The Method of Partial Fractions (3 of 14)

The result is

where S and R are also polynomials.

As the next example illustrates, sometimes this preliminary step is all that is required.

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Example 1

Find

Solution:

Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division.

This enables us to write

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The Method of Partial Fractions (4 of 14)

The next step is to factor the Q(x) as far as possible.

It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax + b) and irreducible quadratic factors (of the form

For instance, if

we could factor it as

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The Method of Partial Fractions (5 of 14)

The third step is to express the proper rational function

(from Equation 1)

as a sum of partial fractions of the form

A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur.

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The Method of Partial Fractions (6 of 14)

Case I The denominator Q(x) is a product of distinct linear factors.

This means that we can write

where no factor is repeated (and no factor is a constant multiple of another).

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The Method of Partial Fractions (7 of 14)

In this case the partial fraction theorem states that there exist constants A₁, A₂, . . . , A_k such that

These constants can be determined as in the following example.

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Example 2

Evaluate

Solution:

Since the degree of the numerator is less than the degree of the denominator, we don’t need to divide.

We factor the denominator as

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Example 2 – Solution (1 of 4)

Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form

To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators, x(2x − 1)(x + 2), obtaining

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Example 2 – Solution (2 of 4)

Expanding the right side of Equation 4 and writing it in the standard form for polynomials, we get

The polynomials on each side of Equation 5 are identical, so their coefficients must be equal.

The coefficient of

on the right side, 2A + B + 2C, must equal the coefficient of

on the left side—namely, 1.

Likewise, the coefficients of x are equal and the constant terms are equal.

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Example 2 – Solution (3 of 4)

This gives the following system of equations for A, B, and C:

Solving, we get,

and so

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Example 2 – Solution (4 of 4)

In integrating the middle term we have made the mental substitution u = 2x − 1,

which gives d u = 2dx and

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The Method of Partial Fractions (8 of 14)

Note:�We can use an alternative method to find the coefficients A, B, and C in Example 2. Equation 4 is an identity; it is true for every value of x. Let’s choose values of x that simplify the equation.

If we put x = 0 in Equation 4, then the second and third terms on the right side

vanish and the equation then becomes −2A = −1, or

Likewise,

gives 10C = −1, so

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The Method of Partial Fractions (9 of 14)

Case II Q(x) is a product of linear factors, some of which are repeated.

Suppose the first linear factor (a₁x + b₁) is repeated r times; that is,

occurs in the factorization of Q(x). Then instead of the single term

in Equation 2, we would use

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The Method of Partial Fractions (10 of 14)

By way of illustration, we could write

but we prefer to work out in detail a simpler example.

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Example 4

Find

Solution:

The first step is to divide. The result of long division is

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Example 4 – Solution (1 of 4)

The second step is to factor the denominator

Since Q(1) = 0, we know that x − 1 is a factor and we obtain

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Example 4 – Solution (2 of 4)

Since the linear factor x − 1 occurs twice, the partial fraction decomposition is

Multiplying by the least common denominator,

we get

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Example 4 – Solution (3 of 4)

Now we equate coefficients:

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Example 4 – Solution (4 of 4)

Solving, we obtain A = 1, B = 2, and C = −1, so

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The Method of Partial Fractions (11 of 14)

Case III Q(x) contains irreducible quadratic factors, none of which is repeated.

If Q(x) has the factor

then, in addition

to the partial fractions in Equations 2 and 7, the expression for

will have a term of the form

where A and B are constants to be determined.

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The Method of Partial Fractions (12 of 14)

For instance, the function given by

has a partial fraction decomposition of the form

The term given in (9) can be integrated by completing the square (if necessary) and using the formula

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Example 6

Evaluate

Solution:

Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain

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Example 6 – Solution (1 of 3)

Notice that the quadratic

is irreducible because its discriminant is

This means it can’t be factored, so we don’t need to use the

partial fraction technique.

To integrate the given function we complete the square in the denominator:

This suggests that we make the substitution u = 2x − 1.

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Example 6 – Solution (2 of 3)

Then d u = 2 dx and

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Example 6 – Solution (3 of 3)

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The Method of Partial Fractions (13 of 14)

Note:�Example 6 illustrates the general procedure for integrating a partial fraction of the form

We complete the square in the denominator and then make a substitution that brings the integral into the form

Then the first integral is a logarithm and the second is expressed in terms of

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The Method of Partial Fractions (14 of 14)

Case IV Q(x) contains a repeated irreducible quadratic factor.

If Q(x) has the factor

then instead of the

single partial fraction (9), the sum

occurs in the partial fraction decomposition of

Each of the terms in (11)

can be integrated by using a substitution or by first completing the square if necessary.

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Example 8

Evaluate

Solution:

The form of the partial fraction decomposition is

Multiplying by

we have

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Example 8 – Solution (1 of 2)

If we equate coefficients, we get the system

which has the solution A = 1, B = −1, C = −1, D = 1 and E = 0.

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Example 8 – Solution (2 of 2)

Thus

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Rationalizing Substitutions

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Rationalizing Substitutions (1 of 1)

Some nonrational functions can be changed into rational functions by means of appropriate substitutions.

In particular, when an integrand contains an expression of the form

then the substitution

may be effective. Other instances appear in the

exercises.

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Example 9

Evaluate

Solution:

Let

Then

and dx = 2u d u. Therefore

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Example 9 – Solution

We can evaluate this integral either by factoring

and using partial fractions or by using Formula 6 with a = 2:

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