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  • Integrate a rational function using the method of partial fractions.
  • Recognize simple linear factors in a rational function.
  • Recognize repeated linear factors in a rational function.
  • Recognize quadratic factors in a rational function.

We have seen some techniques that allow us to integrate specific rational functions. For example, we know that

d u u = ln | u | + C and d u u 2 + a 2 = 1 a tan −1 ( u a ) + C .

However, we do not yet have a technique that allows us to tackle arbitrary quotients of this type. Thus, it is not immediately obvious how to go about evaluating 3 x x 2 x 2 d x . However, we know from material previously developed that

( 1 x + 1 + 2 x 2 ) d x = ln | x + 1 | + 2 ln | x 2 | + C .

In fact, by getting a common denominator, we see that

1 x + 1 + 2 x 2 = 3 x x 2 x 2 .

Consequently,

3 x x 2 x 2 d x = ( 1 x + 1 + 2 x 2 ) d x .

In this section, we examine the method of partial fraction decomposition    , which allows us to decompose rational functions into sums of simpler, more easily integrated rational functions. Using this method, we can rewrite an expression such as: 3 x x 2 x 2 as an expression such as 1 x + 1 + 2 x 2 .

The key to the method of partial fraction decomposition is being able to anticipate the form that the decomposition of a rational function will take. As we shall see, this form is both predictable and highly dependent on the factorization of the denominator of the rational function. It is also extremely important to keep in mind that partial fraction decomposition can be applied to a rational function P ( x ) Q ( x ) only if deg ( P ( x ) ) < deg ( Q ( x ) ) . In the case when deg ( P ( x ) ) deg ( Q ( x ) ) , we must first perform long division to rewrite the quotient P ( x ) Q ( x ) in the form A ( x ) + R ( x ) Q ( x ) , where deg ( R ( x ) ) < deg ( Q ( x ) ) . We then do a partial fraction decomposition on R ( x ) Q ( x ) . The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form P ( x ) Q ( x ) d x , where deg ( P ( x ) ) deg ( Q ( x ) ) .

Integrating P ( x ) Q ( x ) d x , Where deg ( P ( x ) ) deg ( Q ( x ) )

Evaluate x 2 + 3 x + 5 x + 1 d x .

Since deg ( x 2 + 3 x + 5 ) deg ( x + 1 ) , we perform long division to obtain

x 2 + 3 x + 5 x + 1 = x + 2 + 3 x + 1 .

Thus,

x 2 + 3 x + 5 x + 1 d x = ( x + 2 + 3 x + 1 ) d x = 1 2 x 2 + 2 x + 3 ln | x + 1 | + C .
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Visit this website for a review of long division of polynomials.

Evaluate x 3 x + 2 d x .

x 5 ln | x + 2 | + C

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To integrate P ( x ) Q ( x ) d x , where deg ( P ( x ) ) < deg ( Q ( x ) ) , we must begin by factoring Q ( x ) .

Nonrepeated linear factors

If Q ( x ) can be factored as ( a 1 x + b 1 ) ( a 2 x + b 2 ) ( a n x + b n ) , where each linear factor is distinct, then it is possible to find constants A 1 , A 2 ,… A n satisfying

P ( x ) Q ( x ) = A 1 a 1 x + b 1 + A 2 a 2 x + b 2 + + A n a n x + b n .

The proof that such constants exist is beyond the scope of this course.

In this next example, we see how to use partial fractions to integrate a rational function of this type.

Partial fractions with nonrepeated linear factors

Evaluate 3 x + 2 x 3 x 2 2 x d x .

Since deg ( 3 x + 2 ) < deg ( x 3 x 2 2 x ) , we begin by factoring the denominator of 3 x + 2 x 3 x 2 2 x . We can see that x 3 x 2 2 x = x ( x 2 ) ( x + 1 ) . Thus, there are constants A , B , and C satisfying

3 x + 2 x ( x 2 ) ( x + 1 ) = A x + B x 2 + C x + 1 .

We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,

3 x + 2 x ( x 2 ) ( x + 1 ) = A ( x 2 ) ( x + 1 ) + B x ( x + 1 ) + C x ( x 2 ) x ( x 2 ) ( x + 1 ) .

Now, we set the numerators equal to each other, obtaining

3 x + 2 = A ( x 2 ) ( x + 1 ) + B x ( x + 1 ) + C x ( x 2 ) .

There are two different strategies for finding the coefficients A , B , and C . We refer to these as the method of equating coefficients and the method of strategic substitution .

Rule: method of equating coefficients

Rewrite [link] in the form

3 x + 2 = ( A + B + C ) x 2 + ( A + B 2 C ) x + ( −2 A ) .

Equating coefficients produces the system of equations

A + B + C = 0 A + B 2 C = 3 2 A = 2.

To solve this system, we first observe that −2 A = 2 A = −1 . Substituting this value into the first two equations gives us the system

B + C = 1 B 2 C = 2.

Multiplying the second equation by −1 and adding the resulting equation to the first produces

−3 C = 1 ,

which in turn implies that C = 1 3 . Substituting this value into the equation B + C = 1 yields B = 4 3 . Thus, solving these equations yields A = −1 , B = 4 3 , and C = 1 3 .

It is important to note that the system produced by this method is consistent if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.

Rule: method of strategic substitution

The method of strategic substitution is based on the assumption that we have set up the decomposition correctly. If the decomposition is set up correctly, then there must be values of A , B , and C that satisfy [link] for all values of x . That is, this equation must be true for any value of x we care to substitute into it. Therefore, by choosing values of x carefully and substituting them into the equation, we may find A , B , and C easily. For example, if we substitute x = 0 , the equation reduces to 2 = A ( −2 ) ( 1 ) . Solving for A yields A = −1 . Next, by substituting x = 2 , the equation reduces to 8 = B ( 2 ) ( 3 ) , or equivalently B = 4 / 3 . Last, we substitute x = −1 into the equation and obtain −1 = C ( −1 ) ( −3 ) . Solving, we have C = 1 3 .

It is important to keep in mind that if we attempt to use this method with a decomposition that has not been set up correctly, we are still able to find values for the constants, but these constants are meaningless. If we do opt to use the method of strategic substitution, then it is a good idea to check the result by recombining the terms algebraically.

Now that we have the values of A , B , and C , we rewrite the original integral:

3 x + 2 x 3 x 2 2 x d x = ( 1 x + 4 3 · 1 ( x 2 ) 1 3 · 1 ( x + 1 ) ) d x .

Evaluating the integral gives us

3 x + 2 x 3 x 2 2 x d x = ln | x | + 4 3 ln | x 2 | 1 3 ln | x + 1 | + C .
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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