**Introduction to Partial Fractions**

The method of partial fractions is used to integrate rational functions, or quotients of polynomials. We shall treat here some of the basic aspects of the technique.

The first fundamental observation is that there are some elementary rational functions whose integrals we already know.

**I Integrals of Reciprocals of Linear Functions** An integral

with *a* ≠ 0 is always a logarithmic function. In fact we can calculate

**II Integrals of Reciprocals of Quadratic Expressions** An integral

when *a* and *c* are positive, is an inverse trigonometric function. In fact we can use what we learned in Section 6.6.3 to write

**III More Integrals of Reciprocals of Quadratic Expressions** An integral

with *a* > 0, and discriminant *b*^{2} − 4 *ac* negative, will also be an inverse trigonometric function. To see this, we notice that we can write

Since *b*^{2} − 4 *ac* < 0, the final expression in parentheses is positive. For simplicity, let λ = *b* /2 *a* and let *γ = c − b* ^{2} /(4 *a* ). Then our integral is

Of course we can handle this using II above. We find that

**IV Even More on Integrals of Reciprocals of Quadratic Expressions** Evaluation of the integral

when the discriminant *b* ^{2} − 4 *ac* is ≥ 0 will be a consequence of the work we do below with partial fractions. We will say no more about it here.

**Products of Linear Factors**

We illustrate the technique of partial fractions by way of examples.

**Example 1**

Here we treat the case of distinct linear factors.

Let us calculate

**Solution 1**

We notice that the integrand factors as

[Notice that the quadratic polynomial in the denominator will factor *precisely when* the discriminant is ≥ 0, which is case **IV** from Subsection 7.2.1.] Our goal is to write the fraction on the right-hand side of (*) as a sum of simpler fractions. With this thought in mind, we write

where *A* and *B* are constants to be determined. Let us put together the two fractions on the right by placing them over the common denominator ( *x* − 1) × ( *x* − 2). Thus

The only way that the fraction on the far left can equal the fraction on the far right is if their numerators are equal. This observation leads to the equation

1 = *A* ( *x* − 2) + *B* ( *x* − 1)

or

0 = ( *A* + *B* ) *x* + (−2 *A* − *B* − 1).

Now this equation is to be identically true in *x* ; in other words, it must hold for every value of *x* . So the coefficients must be 0.

At long last, then, we have a system of two equations in two unknowns:

*A* + *B* = 0

−2 *A − B* − 1 = 0

Of course this system is easily solved and the solutions found to be *A* = −1, *B* = 1. We conclude that

What we have learned, then, is that

Each of the individual integrals on the right may be evaluated using the information in **I**. As a result,

**You Try It** : Calculate the integral

Now we consider repeated linear factors.

**Example 2**

Let us evaluate the integral

**Solution 2**

In order to apply the method of partial fractions, we first must factor the denominator of the integrand. It is known that every polynomial with real coefficients will factor into linear and quadratic factors. How do we find this factorization? Of course we must find a root. For a polynomial of the form

*x ^{k}* +

*a*

_{k −1}

*x*

^{k −1}+

*a*

_{k −2}

*x*

^{k −2}+ ··· +

*a*

_{1}

*x*+

*a*

_{0},

any integer root will be a factor of *a* _{0} . This leads us to try ±1, ±2, ±3, ±6, ±9 and ±18. We find that −2 and 3 are roots of *x* ^{3} − 4 *x* ^{2} − 3 *x* + 18. In point of fact,

*x*^{3} − 4 *x*^{2} − 3 *x* + 18 = ( *x* + 2) · ( *x* − 3)^{2}.

An attempt to write

*will not work* . We encourage the reader to try this for himself so that he will understand why an extra idea is needed.

In fact we will use the paradigm

Putting the right-hand side over a common denominator yields

Of course the numerators must be equal, so

1 = *A* ( *x* − 3)^{2} + *B* ( *x* + 2)( *x* − 3) + *C* ( *x* + 2).

We rearrange the equation as

(*A* + *B*) *x*^{2} + (−6*A* − *B* + *C*) *x* + (9*A* − 6*B* + 2*C* − 1) = 0.

Since this must be an identity in *x* , we arrive at the system of equations

This system is easily solved to yield *A* = 1/25, *B* = −1/25, *C* = 1/5.

As a result of these calculations, our integral can be transformed as follows:

The first integral equals (1/25) log | *x* + 2|, the second integral equals −(1/25) log | *x* − 3|, and the third integral equals −(1/5)/( *x* − 3).

In summary, we have found that

**You Try It** : Evaluate the integral

** Quadratic Factors Example**

**Example 1**

Evaluate the integral

**Solution 1**

Since the denominator is a cubic polynomial, it must factor. The factors of the constant term are ±1 and ±2. After some experimentation, we find that *x* = −2 is a root and in fact the polynomial factors as

*x* ^{3} + 2 *x* ^{2} + *x* + 2 = ( *x* + 2)( *x* ^{2} + 1).

Thus we wish to write the integrand as the sum of a factor with denominator ( *x* + 2) and another factor with denominator ( *x* ^{2} + 1). The correct way to do this is

We put the right-hand side over a common denominator to obtain

Identifying numerators leads to

*x* = ( *A* + *B* ) *x*^{2} + (2 *B* + *C* ) *x* + ( *A* + 2 *C* ).

This equation must be identically true, so we find (identifying powers of *x*) that

Solving this system, we find that *A* = −2/5, *B* = 2/5, *C* = 1/5. So

**You Try It** : Calculate the integral

**You Try It** : Calculate the integral

Find practice problems and solutions for these concepts at: Methods Of Integration Practice Test.

### Ask a Question

Have questions about this article or topic? Ask### Related Questions

See More Questions### Popular Articles

- Kindergarten Sight Words List
- First Grade Sight Words List
- 10 Fun Activities for Children with Autism
- Grammar Lesson: Complete and Simple Predicates
- Definitions of Social Studies
- Child Development Theories
- Signs Your Child Might Have Asperger's Syndrome
- How to Practice Preschool Letter and Name Writing
- Social Cognitive Theory
- Curriculum Definition