Page 218 - Calculus Demystified

Methods of Integration
CHAPTER 7
Each of the individual integrals on the right may be evaluated using the 205
information in I of Subsection 7.2.1. As a result,

1
dx =− log |x − 1|+ log |x − 2|+ C.
2
x − 3x + 2
You Try It: Calculate the integral
4

dx
.
2
1 x + 5x + 4
Now we consider repeated linear factors.
EXAMPLE 7.6
Let usevaluate the integral

dx
.
2
3
x − 4x − 3x + 18
SOLUTION
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
k
x + 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
3
2
and ±18. We find that −2 and 3 are roots of x − 4x − 3x + 18. In point of
fact,
2
2
3
x − 4x − 3x + 18 = (x + 2) · (x − 3) .
An attempt to write
1 A B
= +
3
2
x − 4x − 3x + 18 x + 2 x − 3
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
1 A B C
= + + .
2
3
x − 4x − 3x + 18 x + 2 x − 3 (x − 3) 2
Putting the right-hand side over a common denominator yields
2
1 A(x − 3) + B(x + 2)(x − 3) + C(x + 2)
= .
2
3
3
2
x − 4x − 3x + 18 x − 4x − 3x + 18