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 = −(3A + 2 B + C);
using the constant coefficient, 4 = 2AThese 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).”
BC ONLY