An Example in Partial Fractions
MATH141
October 20, 2008
How to integrate
x3−2x−4
(x2−x+ 1)(x−1)2.(1)
The factors in the denominator are x−1 with mutiplicity 2, and the irre-
ducible quadratic polynomial x2−x+ 1. Hence the decomposition in partial
fraction is of the form
x3−2x−4
(x2−x+ 1)(x−1)2=A
x−1+B
(x−1)2+Cx +D
x2−x+ 1 (2)
Multiplying both sides of the equation by the denominator on the left hand
side,
x3−2x−4 = A(x−1)(x2−x+ 1) + B(x2−x+ 1)
+Cx(x−1)2+D(x−1)2(3)
=A(x3−2x2+ 2x−1) + B(x2−x+ 1)
+C(x3−2x2+x) + D(x2−2x+ 1)
= (A+C)x3+ (−2A+B−2C+D)x2
+(2A−B+C−2D)x+ (−A+B+D)
Finding A, B, C, D is equivalent to solving the linear system of equations
A+C= 1
−2A+B−2C+D= 0
2A−B+C−2D=−2
−A+B+D=−4
(4)
Another method for finding the coefficients A, B, C, D is to evaluate equation
(3) on the different roots of the denominator of the left-hand side of (1). Since
the only root is 1, this yields
−5 = B
Now take the derivative on both sides of (3) to obtain
1
