# Integration by partial fraction

We shall describe a method for computing the integral of any rational function, and we shall find that the result can always be expressed in terms of polynomials, rational functions, inverse tangents, and logarithms.

The basic idea of the method is to decompose a given rational function into a sum of simpler fractions (called partial fractions) that can be integrated by the techniques discussed earlier. We shall describe the general procedure by means of a number of simple examples that illustrate all the essential features of the method.

EXAMPLE 1. In this example we begin with two simple fractions, and which we know how to integrate, and see what happens when we form a linear combination of these fractions. For example, if we take twice the first fraction plus three times the second, we obtain

If now, we read this formula from right to left, it tells us that the rational function given by has been expressed as a linear combination of and . Therefore, we may evaluate the integral of by writing

EXAMPLE 2. The foregoing example suggests a procedure for dealing with integrals of the form For example, to evaluate , we try to express the integral as a linear combination of and by writing

with constants and to be determined. If we can choose and so that the above equation is an identity, then the integral of the fraction on the left is equal to the sum of the integrals of the simpler fractions on the right. To find and , we multiply both sides by to remove the fractions. This gives us

At this stage there are two methods commonly used to find and . One method is to equate coefficients of like powers of . This leads to the equations and . Solving this pair of simultaneous equations, we obtain and The other method involves the substitution of two values of in the equation and leads to another pair of equations for and . In this particular case, the presence of the factors and suggests that we use the values I and . When we put in the equation, the coefficient of vanishes, and we find , or Similarly, we can make the coefficient of vanish by putting This gives us , or . In any event, we have found values of and to satisfy the equation, so we have

It is clear that the method described in Example 2 also applies to integrals of the form in which is a linear polynomial and is a quadratic polynomial that can be factored into distinct linear factors with real coefficients, say In this case the quotient can be expressed as a linear combination of and , and integration of leads to a corresponding combination of the logarithmic terms and .

The foregoing examples involve rational functions in which the degree of the numerator is less than that of the denominator. A rational function with this property is said to be a proper rational function. If is improper, that is, if the degree of is not less than that of , then we can express as the sum of a polynomial and a proper rational function. In fact, we simply divided by to obtain

where and are polynomials (called the quotient and remainder, respectively) such that the remainder has degree less than that of . For example,

Therefore, in the study of integration technique, there is no loss in generality if we restrict ourselves to proper rational functions, and from now on we consider , where has degree less than that of .

A general theorem in algebra states that every proper rational function can be expressed as a finite sum of fractions of the forms

where and are positive integers and are constants with The condition means that the quadratic polynomial cannot be factored into linear factors with real coefficients or, what amounts to the same thing, the quadratic equation has no real roots. Such a quadratic factor is said to be irreducible. When a rational function has been so expressed, we say that it has been decomposed into partial fractions. Therefore the problem of integrating this rational function reduces to that of integrating its partial fractions. These may be easily dealt with by the techniques described in the examples which follow.

We shall not bother to prove that partial-fraction decompositions always exist. Instead, we shall show (by means of examples) how to obtain the partial fractions in specific problems. In each case that arises the partial-fraction decomposition can be verified directly.

It is convenient to separate the discussion into cases depending on the way in which the denominator of the quotient can be factored.

CASE 1. The denominator is a product of distinct linear factors. Suppose that splits into distinct linear factors, say

Now notice that a linear combination of the form

may be expressed as a single fraction with the common denominator and the numerator of this fraction will be a polynomial of degree involving the 's. Therefore, if we can find 's to make this numerator equal to , we shall have the decomposition

and the integral of will be equal to In the next example, we work out a case with

EXAMPLE 3. Integrate

Solution. Since , the denominator is a product of distinct linear factors, and we try to find and such that

Clearing the fractions, we obtain

When we find so When we obtain , and when we find or Therefore we have

CASE 2. The denominator is a product of linear factors, some of which are repeated. We illustrate this case with an example.

EXAMPLE 4. Integrate .

Solution. Here we try to find so that

We need both and as well as in order to get a polynomial of degree two in the numerator and to have as many constants as equations when we try to determine the 's. Clearing the fractions, we obtain

Substituting , we find , so . When we obtain and We need one more equation to determine Since there are no other choices of that will make any factor vanish, we choose a convenient that will help to simplify the calculations. For example, the choice leads to the equation from which we find An alternative method is to differentiate both sides of the equation and then substitute a convenient x. Differentiation leads to the equation

and, if we put we find so as before. Therefore we have found 's to satisfy the equation, so we have

If, on the left of the equation the factor had appeared instead of , we would have added an extra term on the right. More generally, if a linear factor appears times in the denominator, then for this factor we must allow for a sum of terms, namely

where the 's are constants. A sum of this type is to be used for each repeated linear factor.

CASE 3. The denominator contains irreducible quadratic factors, none of which are repeated.

EXAMPLE 5. Integrate .

Solution. The denominator can be split as the product , where is irreducible, and we try a decomposition of the form

In the fraction with denominator , we have used a linear polynomial in the numerator in order to have as many constants as equations when we solve for Clearing the fractions and solving for and we find and Therefore we have

The first integral on the right is . To evaluate the second integral, we write

If we let and , the last integral is

Therefore, we have

CASE 4. The denominator contains irreducible quadratic factors, some of which are repeated. Here the situation is analogous to Case 2. In the partial fraction decomposition of we allow first of all, a sum of the form

for each linear factor, as already described. In addition. if an irreducible quadratic factor is repeated times, we allow a sum of terms, namely

where each numerator is linear.

EXAMPLE 6. Integrate

Solution. We write

Clearing the fractions and solving for and we find that

Therefore, we have

The foregoing examples are typical of what happens in general. The problem of integrating a proper rational function reduces to that of calculating integrals of the forms

The first integral is if and if To treat the other two, we express the quadratic as a sum of two squares by writing

where and . (This is possible because .) The substitution reduces the problem to that of computing

The first of these is if , and if When , the second integral is evaluated by the formula

The case may be reduced to the case by repeated application of the recursion formula

which is obtained by integration by parts. This discussion shows that every rational function may he integrated in terms of polynomials, rational functions, inverse tangents, and logarithms.

# Integrals which can be transformed into integrals of rational functions

A function of two variables defined by an equation of the form

is called a polynomial in two variables. The quotient of two such polynomials is called a rational function of two variables. Integrals of the form where is a rational function of two variables, may be reduced by the substitution to integrals of the form where is a rational function of one variable. The latter integral may be evaluated by the techniques just described. We illustrate the method with a particular example.

EXAMPLE 1. Integrate .

Solution. The substitution gives us

and

Therefore, we have

where and . The method of partial fractions leads to

and, since , we obtain

The final answer may be simplified somewhat by using suitable trigonometric identities. First we note that so the numerator of the last fraction is In the denominator we write

We may combine the term with the arbitrary constant and rewrite as follows:

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