integration tutorial: analysis

Consider the integral
int (q + 11)/(2q - 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:
  1. Recognize elementary antiderivatives
  2. Rewrite the integrand to make it easier
  3. Use substitution to reverse the chain rule or simplify the integrand
  4. Use integration by parts
  5. 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 = 2q - 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 = 2q - 6. Then w' = 2, so (1/2)dw = dq. The integral can therefore be rewritten as
int (q + 11)/(2q - 6) dq = int (1/2) + 14/w (1/2)dw = (1/4) w + 7 ln(|w|) + C
Thus, substituting back for w,
int (q + 11)/(2q - 6) dq = (1/4) (2q - 6) + 7 ln(|2q - 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,
int (q + 11)/(2q - 6) dq =int1/2dq+int7/(q - 3)dq
which can be evaluated using basic antidifferentiation, and substitution, respectively, to obtain
int1/2dq+int7/(q - 3)dq = ((1/2)q) + (7 ln(abs(q - 3))) + C.

Explanation for rewritten term(s)
Basic antidifferentiation
In this case, by using rules for basic antiderivatives we can just write down the antiderivative of the integrand:
int 1/2 dq = (1/2)q + C

Substitution
Let w = q - 3. Then w' = 1, so dw = dq. The integral can therefore be rewritten as

int 7/(q - 3) dq = int 7/w dw = 7 ln(abs(w)) + C
Thus, substituting back for w,
int 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.