Page 216 - Calculus Demystified

Methods of Integration
CHAPTER 7
2
with a> 0, and discriminant b − 4ac negative, will also be an inverse trigono- 203
metric function. To see this, we notice that we can write

b
2 2
ax + bx + c = a x + x + + c
a
2 2
2 b b b
= a x + x + + c −
a 4a 2 4a
2 2
b b
= a · x + + c − .
2a 4a
2
Since b − 4ac < 0, the final expression in parentheses is positive. For simplicity,
2
let λ = b/2a and let γ = c − b /(4a). Then our integral is

1
dx.
γ + a · (x + λ) 2
Of course we can handle this using II above. We find that

1 1
dx = dx
2
ax + bx + c γ + a · (x + λ) 2
√
1 a
= √ · arctan √ · (x + λ) + C.
aγ γ
IV Even More on Integrals of Reciprocals of Quadratic Expressions
Evaluation of the integral

1
dx
2
ax + bx + c
2
when the discriminant b − 4ac is ≥ 0 will be a consequence of the work we do
below with partial fractions. We will say no more about it here.

7.2.2 PRODUCTS OF LINEAR FACTORS

We illustrate the technique of partial fractions by way of examples.

EXAMPLE 7.5
Here we treat the case of distinct linear factors.
Let uscalculate

1
dx.
2
x − 3x + 2