21B INTEGRATION
The first step in all integration problems is to determine which type of problem it is…
algebra
(no exponentials
(e's), logs or trig),
trig
or
mixture
(some combination of algebra, exponentials (e's), logs and trig). The following outline
suggests an orderly approach to deciding which integration technique to use.
Algebra
Trig
Mixture
* Can you make it an
exponents problem?
* Does the derivative of one
part = the other part?
* Does the derivative of one
part = the other part?
* Does the derivative of one
part equal the other part?
* Tricks…
* Integration by Parts
(LATE)
* Tricks…
Long division
Partial Fractions
Complete the square
Trig substitution
Algebraic
Substitution
I. Algebra
A. Can you make it an exponent problem?
[Any necessary quantities in parentheses are raised to a positive whole number
power.]
By an exponent problem, I mean any expression which may be written as the sums and/or differences of terms of
the form ax
n
where a and n are constants. [e.g.
10
x
1 2
!
3
x
!
2
+
4
x
!
2 3
]
For example:
(2x+3)
2
=
4
x
2
+
12
x
+
9
3
x
2
+
1
2
x
2
=
3
x
2
+
1
( )
!
1
2
x
"
2
=
3
2
+
1
2
x
"
2
are each written in the desired form. Note that any terms in parentheses are raised to a whole number power.
These functions may now be integrated by the power rule.
3
x
2
+
1
2
x
2
!
dx
=
3
2
+
1
2
x
"
2
!
dx
=
3
2
x
"
1
2
x
"
1
+
C
Integrands like
3
x
+
1
=
(3
x
+
1)
1 2
and
1
(
x
2
+
3)
2
=
x
2
+
3
( )
!
2
require parentheses with powers that are not
positive whole numbers and therefore cannot be written as exponent problems.
B. Does the derivative of one part = the other part? [
Substitution  The degree of one part is one more than the degree of
the other part. Let u = the part with the higher degree.]
Taking the derivative of an expression lowers the degree of the expression by one. Therefore, if the derivative of
one part is to equal the other part, it is clear that the powers must differ by one.
For example the degree of 4x
2
+ 2x +13 and that of 8x + 2 differ by one (2 and 1 respectively) so in this case try
letting u = 4x
2
+ 2x  13, the one with the larger power. Since
du
= (8x + 2)
dx
we are in luck.
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x
+
2
(4
x
2
+
2
x
+
13)
3
!
dx
=
du
u
3
!
= so.
..
u
!
3
"
du
=
!
1
2
u
!
2
+
C
=
!
1
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 Spring '06
 Sarason
 Calculus, Algebra, Trigonometry, dx, Mathematics in medieval Islam, exa mple

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