INTEGRATION BY PARTIAL FRACTIONS - Antidifferentiation - Calculus AB and Calculus BC

Calculus AB and Calculus BC

CHAPTER 5 Antidifferentiation

C. INTEGRATION BY PARTIAL FRACTIONS

The method of partial fractions makes it possible to express a rational function Image as a sum of simpler fractions. Here f (x) and g(x) are real polynomials in x and it is assumed that Image is a proper fraction; that is, that f (x) is of lower degree than g(x). If not, we divide f (x) by g(x) to express the given rational function as the sum of a polynomial and a proper rational function. Thus,

Image

where the fraction on the right is proper.

Theoretically, every real polynomial can be expressed as a product of (powers of) real linear factors and (powers of) real quadratic factors.

In the following, the capital letters denote constants to be determined. We consider only nonrepeating linear factors. For each distinct linear factor (x a) of g(x) we set up one partial fraction of the type Image The techniques for determining the unknown constants are illustrated in the following examples.

Examples 42–47 are BC ONLY.

EXAMPLE 42

Find Image

SOLUTION: We factor the denominator and then set

Image

where the constants A, B, and C are to be determined. It follows that

Image

Since the polynomial on the right in (2) is to be identical to the one on the left, we can find the constants by either of the following methods:

METHOD ONE. We expand and combine the terms on the right in (2), getting

x2 x + 4 = (A + B + C)x2 − (3A + 2B + C)x + 2A.

We then equate coefficients of like powers in x and solve simultaneously. Thus

using the coefficients of x2, we get

1 = A + B + C;

using the coefficients of x, we get

−1 = −(3A + 2B + C);

using the constant coefficient,

4 = 2A

These equations yield A = 2, B = −4, C = 3.

METHOD TWO. Although equation (1) above is meaningless for x = 0, x = 1, or x = 2, it is still true that equation (2) must hold even for these special values. We see, in (2), that

if x = 0,

then 4 = 2A and A = 2;

if x = 1,

then 4 = −B and B = −4;

if x = 2,

then 6 = 2C and C = 3.

The second method is shorter than the first and more convenient when the denominator of the given fraction can be decomposed into nonrepeating linear factors.

Finally, then, the original integral equals

Image

[The symbol “C′” appears here for the constant of integration because C was used in simplifying the original rational function.]

In the Topical Outline for Calculus BC, integration by partial fractions is restricted to “simple partial fractions (nonrepeating linear factors only).”