Page 219 - Calculus Demystified

CHAPTER 7
Methods of Integration
206
Of course the numerators must be equal, so
2
1 = A(x − 3) + B(x + 2)(x − 3) + C(x + 2).
We rearrange the equation as
2
(A + B)x + (−6A − B + C)x + (9A − 6B + 2C − 1) = 0.
Since this must be an identity in x, we arrive at the system of equations

A + B = 0
−6A − B + C = 0
9A − 6B + 2C − 1 = 0
This system is easily solved to yield A = 1/25, B =−1/25, C = 1/5.
As a result of these calculations, our integral can be transformed as follows:

1 1/25 −1/25 1/5
dx = dx + dx + dx.
2
3
x − 4x − 3x + 18 x + 2 x − 3 (x − 3) 2
The first integral equals (1/25) log |x + 2|, the second integral equals −(1/25)
log |x − 3|, and the third integral equals −(1/5)/(x − 3).
In summary, we have found that

1 log |x + 2| log |x − 3| 1
dx = − − + C.
3
2
x − 4x − 3x + 18 25 25 5(x − 3)
You Try It: Evaluate the integral

4
xdx
.
3
2
2 x + 5x + 7x + 3
7.2.3 QUADRATIC FACTORS
EXAMPLE 7.7
Evaluate the integral

xdx
.
3
2
x + 2x + x + 2
SOLUTION
Since the denominator is a cubic polynomial, it must factor. The factors of
the constant term are ±1 and ±2. After some experimentation, we find that
x =−2 is a root and in fact the polynomial factors as
2
3
2
x + 2x + x + 2 = (x + 2)(x + 1).