Barrons AP Calculus - David Bock

Find

SOLUTION: We factor the denominator and then set

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

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

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

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

using the coefficients of x^2 , we get 1 = A + B + C;

using the coefficients of x, we get −1 = −(3C); A + 2B +

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

[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).”