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## 7.4 Integration of Rational Functions

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### Integration of Rational Functions by Partial Fractions

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 2/(x – 1) and 1/(x + 2) to a common denominator we obtain

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### Integration of Rational Functions by Partial Fractions

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

= 2 ln |x – 1 | – ln | x + 2 | + C

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### Integration of Rational Functions by Partial Fractions

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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### Integration of Rational Functions by Partial Fractions

We know that if

P(x) = anxn + an – 1xn – 1 + . . . + a1x + a0

where an ≠ 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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### Integration of Rational Functions by Partial Fractions

The division statement 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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### Integration of Rational Functions by Partial Fractions

The next step is to factor the denominator 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 ax2 + bx + c, where

b2 – 4ac < 0).

For instance, if Q(x) = x4 – 16, we could factor it as Q(x) = (x2 – 4)(x2 + 4) = (x – 2)(x + 2)(x2 + 4)

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### Integration of Rational Functions by Partial Fractions

The third step is to express the proper rational function R(x)/Q(x) (from Equation 1) as a sum of partial fractions of the form

or

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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### Integration of Rational Functions by Partial Fractions

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

This means that we can write

Q(x) = (a1x + b1)(a2x + b2) . . . (akx + bk)

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

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### Integration of Rational Functions by Partial Fractions

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

2x3 + 3x2 – 2x = x(2x2 + 3x – 2)

= x(2x – 1)(x + 2)

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### Example 2 – Solution

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

x2 + 2x – 1 = A(2x – 1)(x + 2) + Bx(x + 2) + Cx(2x – 1)

cont’d

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### Example 2 – Solution

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

x2 + 2x – 1 = (2A + B + 2C)x2 + (3A + 2B – C)x – 2A The polynomials in Equation 5 are identical, so their

coefficients must be equal. The coefficient of x2 on the right side, 2A + B + 2C, must equal the coefficient of x2 on the left side—namely, 1.

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

cont’d

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### Example 2 – Solution

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

2A + B + 2C = 1 3A + 2B – C = 2 –2A = –1

Solving, we get, A = B = and C = and so

cont’d

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### Example 2 – Solution

In integrating the middle term we have made the mental substitution u = 2x – 1, which gives du = 2dx and

dx = du.

cont’d

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### Integration of Rational Functions by Partial Fractions

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 A = .

Likewise, x = gives 5B/4 = and x = –2 gives 10C = –1, so

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### Integration of Rational Functions by Partial Fractions

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, (a1x + b1)r occurs in the factorization of Q(x). Then instead of the single term A1/(a1x + b1) in Equation 2, we would use

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### Integration of Rational Functions by Partial Fractions

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

The second step is to factor the denominator Q(x) = x3 – x2 – x + 1.

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

x3 – x2 – x + 1 = (x – 1)(x2 – 1)

= (x – 1)(x – 1)(x + 1)

= (x – 1)2(x + 1)

cont’d

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### Example 4 – Solution

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

Multiplying by the least common denominator, (x – 1)2(x + 1), we get

4x = A(x – 1)(x + 1) + B(x + 1) + C(x – 1)2

cont’d

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### Example 4 – Solution

= (A + C)x2 + (B – 2C)x + (–A + B + C) Now we equate coefficients:

A + C = 0 B – 2C = 4 –A + B + C = 0

cont’d

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### Example 4 – Solution

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

cont’d

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### Integration of Rational Functions by Partial Fractions

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

If Q(x) has the factor ax2 + bx + c, where b2 – 4ac < 0, then, in addition to the partial fractions in Equations 2 and 7, the expression for R(x)/Q(x) will have a term of the form

where A and B are constants to be determined.

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### Integration of Rational Functions by Partial Fractions

For instance, the function given by

f(x) = x/[(x – 2)(x2 + 1)(x2 + 4)] 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

Notice that the quadratic 4x2 – 4x + 3 is irreducible because its discriminant is b2 – 4ac = –32 < 0. 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:

4x2 – 4x + 3 = (2x – 1)2 + 2

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

cont’d

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### Example 6 – Solution

Then du = 2 dx and x = (u + 1), so

cont’d

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cont’d

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### Integration of Rational Functions by Partial Fractions

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

where b2 – 4ac < 0

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### Integration of Rational Functions by Partial Fractions

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

If Q(x) has the factor (ax2 + bx + c)r, where b2 – 4ac < 0, then instead of the single partial fraction (9), the sum

occurs in the partial fraction decomposition of R(x)/Q(x).

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 x(x2 + 1)2, we have

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### Example 8 – Solution

= A(x4 + 2x2 + 1) + B(x4 + x2) + C(x3 + x) + Dx2 + Ex

= (A + B)x4 + Cx3 + (2A + B + D)x2 + (C + E)x + A If we equate coefficients, we get the system

A + B = 0 C = –1 2A + B + D = 2 C + E = –1 A = 1

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

cont’d

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Thus

cont’d

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### Rationalizing Substitutions

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 u = Then u2 = x + 4, so x = u2 – 4 and dx = 2u du. Therefore

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### Example 9 – Solution

We can evaluate this integral either by factoring u2 – 4 as (u – 2)(u + 2) and using partial fractions or by using

Formula 6 with a = 2:

cont’d

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