文档介绍：Section
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Integration of Rational Functions
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In this section we will take a more detailed look at the use of partial fraction posi-
tions in evaluating integrals of rational functions, a technique we first encountered in the
inhibited growth model example in the previous section. However, we will not be able to
complete the story until after the introduction of the inverse tangent function in Section
.
We begin with a few examples to illustrate how some integration problems involving
rational functions may be simplified either by a long division or by a simple substitution.
x2
Example To evaluate dx, we first perform a long division of x + 1 into x2 to
Z x + 1
obtain
x2 1
= x − 1 + .
x + 1 x + 1
Then
x2 1 1
dx = x − 1 + dx = x2 − x + log |x + 1| + c.
Z x + 1 Z x + 1 2
2x + 1
Example To evaluate dx, we make the substitution
Z x2 + x
u = x2 + x
du = (2x + 1)dx.
Then
2x + 1 1
dx = du = log |u| + c = log |x2 + x| + c.
Z x2 + x Z u
x
Example To evaluate dx, we perform a long division of x + 1 into x to obtain
Z x + 1
x 1
= 1 −.
x + 1 x + 1
Then
x 1
dx = 1 − dx = x − log |x + 1| + c.
Z x + 1 Z x + 1
1
2 Integration of Rational Functions Section
Alternatively, we could evaluate this integral with the substitution
u = x + 1