integration tutorial: analysis
Consider the integral

(
q + 11)/(2
q - 6)
dq
This is an indefinite integral,
so we need to find
the most general
antiderivative of the integrand f(q).
To find an antiderivative of
f(
q), we go through
our list of integration methods:
- Recognize elementary antiderivatives
- Rewrite the integrand to make it easier
- Use substitution to reverse the chain rule or simplify the
integrand
- Use integration by parts
- Use inspection to see the value of a definite
integral
to find one that works for this integrand. In this case,
there are several methods that work. We can use substitution
to rewrite the integrand in terms of
w = 2
q - 6, or can rewrite the integrand: use long division to divide the denominator into the
numerator to give an expression that is more easily integrable.
This is shown below:
substitution;
rewriting.
Substitution:
Let
w = 2
q - 6. Then
w' = 2, so
(1/2)
dw =
dq.
The integral can therefore be rewritten as

(
q + 11)/(2
q - 6)
dq =

(1/2) + 14/
w (1/2)
dw =
(1/4)
w + 7 ln(|
w|) +
C
Thus, substituting back for
w,

(
q + 11)/(2
q - 6)
dq =
(1/4) (2
q - 6) + 7 ln(|2
q - 6|) +
C
Rewriting:
Use long division to divide the denominator into the
numerator to rewrite the integral:
(q + 11)/(2q - 6)
=
(1/2) + 7/(q - 3).
This can be broken into separate integrals,

(
q + 11)/(2
q - 6)
dq
=

1/2
dq+

7/(
q - 3)
dq
which can be evaluated using
basic antidifferentiation, and
substitution,
respectively, to obtain

1/2
dq+

7/(
q - 3)
dq
=
((1/2)
q) + (7 ln(abs(
q - 3))) +
C.
Explanation for rewritten term(s)
Basic antidifferentiationIn this case, by using
rules for basic
antiderivatives we can just write down the antiderivative of the
integrand:

1/2
dq
=
(1/2)
q +
C
SubstitutionLet
w =
q - 3. Then
w' = 1, so
dw =
dq.
The integral can therefore be rewritten as

7/(
q - 3)
dq =

7/
w dw =
7 ln(abs(
w)) +
C
Thus, substituting back for
w,

7/(
q - 3)
dq =
7 ln(abs(
q - 3)) +
C
Note that when there are multiple integration
methods the antiderivatives may appear to be different but are
actually the same function shifted vertically by different constants.
This is reasonable because the most general antiderivative is only
unique up to an additive constant.
integration analysis
Last Modified: Wed Feb 6 13:53:59 EST 2002
Comments to
glarose@umich.edu
©2002 Gavin LaRose, UM Math Dept.