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Unit 3:
Integration Techniques
We have learned how to integrate functions of the following friendly forms,
and have learned how to use the method of substitution to transform some
problems into these forms:
i
u
n
du
=
u
n
+1
n
+1
+
C,
n
n
=

1
i
1
u
du
= ln

u

+
C
i
e
u
du
=
e
u
+
C
But sometimes this is not enough. For instance, if we wish to evaluate
something like
i
x
ln(
x
)
dx
, nothing we have learned so far will work. For
something like this we turn to a new method, namely that of
integration by
parts
, which is sometimes called “double substitution”.
The derivation of the formula is easily seen to be a consequence of the
product rule. When
u
and
v
are functions of
x
, we have
d
dx
[
uv
] =
uv
p
+
vu
p
where, of course,
v
p
=
dv
dx
and
u
p
=
du
dx
. Antidi±erentiating both sides of this
equation with respect to
x
we get
uv
=
I
uv
p
dx
+
I
vu
p
dx
which can be rearranged to
I
uv
p
dx
=
uv

I
vu
p
dx
This is the formula we use. Notice that since
v
p
=
dv
dx
, then
v
p
dx
=
dv
dx
dx
=
dv
and since
u
p
=
du
dx
, then
u
p
dx
=
du
dx
dx
=
du
. We get:
1
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View Full Document Defnition 3.1.
The
integration by parts
formula:
If
u
and
v
are diFerentiable functions then,
i
udv
=
uv

i
vdu
This formula is useful when
i
vdu
is an integral which is easier to evaluate
than the one we started with,
i
udv
. So when facing a di±cult integral, we
consider whether we can recognize the form
i
udv
for some
u
and
v
.
The following steps may make it a bit easier to use this method.
Step 1:
Write the given integral in the form
i
f
(
x
)
g
(
x
)
dx
so that you can
identify the two functions
f
(
x
) and
g
(
x
). Note that if there appears to be
only one function in the integrand, then you may need to identify the second
one to be the constant function,
g
(
x
) = 1.
Step 2:
Introduce the functions
u
(
x
) and
v
(
x
) by letting:
u
=
f
(
x
)
and
dv
=
g
(
x
)
dx
Now, to use the integration by parts formula, we need to ²nd
du
and
v
. That
is, we diFerentiate
u
to get:
du
dx
=
f
p
(
x
)
→
du
=
f
p
(
x
)
dx
and we integrate
dv
to get:
v
=
i
dv
→
v
=
i
g
(
x
)
dx
Step 3:
Use the formula to reexpress the original integral:
replace
i
udv
with
uv

i
vdu
Step 4:
Evaluate the new expression (by ²nding
i
vdu
).
Note:
When we ²nd
v
=
i
dv
=
i
v
p
dx
in Step 2, we don’t need to include
an arbitrary constant. This is because when we use
v
in our formula, we will
still have an inde²nite integral appearing in the expression. All arbitrary
constants will be collected together as one when we perform the last integra
tion.
2
Of course, the main diFculty in this method is determining the choice of
u
and
dv
. In Step 1, we recognize the integrand as the product of 2 terms.
But how, in Step 2, do we determine which term should be
u
and which
should be
dv
? There is no general rule to follow. It is truly a matter of
experience. However, there are 2 guiding principles you should keep in mind:
(1)
dv
must be something you know how to integrate, and (2) we want
i
vdu
to be less complicated than
i
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This note was uploaded on 11/03/2009 for the course MATH 0110A taught by Professor Olds,vicky during the Fall '09 term at UWO.
 Fall '09
 OLDS,VICKY
 Calculus

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