Page 217 - Calculus Demystified

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SOLUTION CHAPTER 7 Methods of Integration
We notice that the integrand factors as
1 1
= . (∗)
2
x − 3x + 2 (x − 1)(x − 2)
[Notice that the quadratic polynomial in the denominator will factor precisely
when the discriminant is ≥ 0, which is case IV from Subsection 7.2.1.] Our
goal is to write the fraction on the right-hand side of (∗) as a sum of simpler
fractions. With this thought in mind, we write

1 A B
= + ,
(x − 1)(x − 2) x − 1 x − 2
where A and B are constants to be determined. Let us put together the two
fractions on the right by placing them over the common denominator (x −1)×
(x − 2). Thus

1 A B A(x − 2) + B(x − 1)
= + = .
(x − 1)(x − 2) x − 1 x − 2 (x − 1)(x − 2)
The only way that the fraction on the far left can equal the fraction on the far
right is if their numerators are equal. This observation leads to the equation
1 = A(x − 2) + B(x − 1)

or
0 = (A + B)x + (−2A − B − 1).

Now this equation is to be identically true in x; in other words, it must hold for
every value of x. So the coefficients must be 0.
At long last, then, we have a system of two equations in two unknowns:

A + B = 0
−2A − B − 1 = 0

Of course this system is easily solved and the solutions found to be A =
−1,B = 1. We conclude that

1 −1 1
= + .
(x − 1)(x − 2) x − 1 x − 2
What we have learned, then, is that

1 −1 1
dx = dx + dx.
2
x − 3x + 2 x − 1 x − 2