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Partial Fractions Help

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By — McGraw-Hill Professional
Updated on Aug 31, 2011

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

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

II Integrals of Reciprocals of Quadratic Expressions An integral

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

III More Integrals of Reciprocals of Quadratic Expressions An integral

Methods of Integration 7.2 Partial Fractions

with a > 0, and discriminant b2 − 4 ac negative, will also be an inverse trigonometric function. To see this, we notice that we can write

Methods of Integration 7.2 Partial Fractions

Since b2 − 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

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

Solution 1

We notice that the integrand factors as

Methods of Integration 7.2 Partial Fractions

[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

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

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 AB − 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

Methods of Integration 7.2 Partial Fractions

What we have learned, then, is that

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

You Try It : Calculate the integral

Methods of Integration 7.2 Partial Fractions

Now we consider repeated linear factors.

Example 2

Let us evaluate the integral

Methods of Integration 7.2 Partial Fractions

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,

x3 − 4 x2 − 3 x + 18 = ( x + 2) · ( x − 3)2.

An attempt to write

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

Putting the right-hand side over a common denominator yields

Methods of Integration 7.2 Partial Fractions

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) x2 + (−6AB + C) x + (9A − 6B + 2C − 1) = 0.

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

Methods of Integration 7.2 Partial Fractions

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:

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

You Try It : Evaluate the integral

Methods of Integration 7.2 Partial Fractions

Quadratic Factors Example

Example 1

Evaluate the integral

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

Identifying numerators leads to

x = ( A + B ) x2 + (2 B + C ) x + ( A + 2 C ).

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

Methods of Integration 7.2 Partial Fractions

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

Methods of Integration 7.2 Partial Fractions

You Try It : Calculate the integral

Methods of Integration 7.2 Partial Fractions

You Try It : Calculate the integral

Methods of Integration 7.2 Partial Fractions

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

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