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:
R
un du =
un+1
n+1
+ C,
R
1
du
u
R
eu du = eu + C
n 6= −1
= ln |u| + C
But sometimesR this is not enough. For instance, if we wish to evaluate
something like 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
[uv] = uv 0 + vu0
dx
dv
0
where, of course, v = dx and u0 = du
. Antidifferentiating both sides of this
dx
equation with respect to x we get
Z
Z
0
uv = uv dx + vu0 dx
which can be rearranged to
Z
0
uv dx = uv −
Z
vu0 dx
dv
, then v 0 dx =
This is the formula we use. Notice that since v 0 = dx
du
du
0
0
and since u = dx , then u dx = dx dx = du. We get:
1
dv
dx
dx
= dv
Definition 3.1. The integration by parts formula:
If u and v are differentiable functions then,
R
udv = uv −
R
vdu
R
This formula is useful when Rvdu is an integral which is easier to evaluate
than the one we started with, udv. So when
R facing a difficult integral, we
consider whether we can recognize the form udv for some u and v.
The following steps may make it a bit easier toR use this method.
Step 1: Write the given integral in the form 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 find du and v. That
is, we differentiate u to get:
du
dx
= f 0 (x) → du = f 0 (x)dx
and we integrate dv to get:
v=
R
dv → v =
R
g(x)dx
Step 3: Use the formula to re-express the original integral:
R
R
replace udv with uv − vdu
R
Step 4: Evaluate the new expression (by finding vdu).
R
R
Note: When we find v = dv = v 0 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 indefinite integral appearing in the expression. All arbitrary
constants will be collected together as one when we perform the last integration.
2
Of course, the main difficulty 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:
R
(1) dv must be something youR know how to integrate, and (2) we want vdu
to be less complicated than udv, i.e. than what we started with. In general it is safest to pick dv to be the “most complicated” choice that can be
integrated easily. In order to determine whether a choice is the best one, go
on until you get to Step 4. If the new integral is easier than the original one,
then your choice is a good one. Otherwise go back to Step 2 and make a
switch. With practice, you will start to have a feeling for the right choice.
Example 1. Find
R
x ln xdx.
Solution: Looking at the integrand, we don’t recognize x ln x as the derivative of any function. Also, we can quickly determine that no substitution
will work here. (You should check that for yourself.) We turn to integration
by parts.
We have a product of 2 functions, x and ln x. In deciding which should be
u and which should be v 0 , we need to realize that we don’t know how to
integrate ln x. However, we do know how to differentiate it. Thus, we should
choose this as the function to be differentiated. That is, we let:
u = ln(x) and dv = xdx
d
This gives du
= dx
(ln x) = x1 , so that du = x1 dx, and also v =
dx
Using the integration by parts formula, we see that
Z
Z
udv = uv − vdu
becomes
R
x ln xdx
=
2
(ln x)( x2 )
=
x2
2
=
x2 ln x
2
R
x
dx
2
x2
4
+C
ln x −
−
3
−
R x2 2
1
x
dx
R
xdx =
x2
.
2
Example 2. Find
R
xe3x dx.
Solution: Again, we see that this is an integral we cannot evaluate by other
means. (You should confirm that there is not a substitution that will work
here.) Thus, we use integration by parts.
Notice: The function f (x) = x gets less complicated when we differentiate
(f 0 (x) = 1) but gets more complicated when we integrate (an antiderivative of
2
x is x2 ). However, g(x) = e3x gets neither more nor less complicated, whether
we differentiate or integrate (i.e g 0 (x) = 3e3x R, while an antiderivative of g(x)
1 3x
Ris 3 e .) Because of this, in order to have vdu be less complicated than
udv, we make the choice that makes f (x) = x get less complicated:
u = x and dv = e3x dx
3x
This gives du = dx and v = e3 . Using the integration by parts formula, we
see that
Z
Z
udv = uv − vdu
becomes
R
Example 3. Find
R
3x
xe3x dx = x e3 −
=
xe3x
3
R
e3x
dx
3
−
1
3
R
e3x dx
=
xe3x
3
−
1
3
e3x
3
=
e3x
3
x−
1
3
+C
+C
ln xdx.
Solution: Off hand, we don’t know any function whose derivative is ln x.
As well, there doesn’t appear to be any way to make a substitution. This
suggests that we should try to use integration by parts − but how? The
integrand does not seem to be a product of 2 functions.
Or is it? We can consider ln x as (ln x)(1). Then letting u = ln x and
dv = 1dx, we get du = x1 dx and v = x.
4
We get:
Z
ln xdx =
Z
udv = uv −
= x ln x −
Example 4. Find
R
Z
Z
vdu = (ln x)(x) −
Z
1
x
dx
x
1dx = x ln x − x + C
x2 ex dx.
Solution: We can easily establish that this is not an integral we recognize,
and that the substitution rule does not help. (You should check that for
yourself.) So integration by parts would seem to be the correct approach.
Considering
the R2 functions f (x) = x2 and g(x) = ex , we once again see
R
3
that f (x)dx = x2 dx = x3 (+C) is a more complicated function than f (x)
(i.e., higher
powerRof x), while f 0 (x) = 2x is less complicated. On the other
R
hand, g(x)dx = ex dx = ex (+C) and f 0 (x) = ex , so this function does not
get more complicated no matter which way we go. With the goal of making
things less complicated, we therefore choose u = x2 and dv = ex dx. This
gives du = 2xdx and v = ex , so we get
Z
Z
Z
Z
2 x
2
x
x e dx = udv = uv − vdu = (x )(e ) − (ex ) (2x)dx
2 x
=x e −2
Z
xex dx
R
We see that xex dx is less complicated, but is still not something we know
(and the substitution rule does not help). What to do? Use integration by
parts again.
This time, the integrand is xex . Using the same reasoning as before, we let
u = x and dv = ex dx, so that du = dx and v = ex . Then we have
Z
Z
Z
Z
x
x
xe dx = udv = uv − vdu = xe − ex dx = xex − ex + C
Going back to our original problem, we have
Z
Z
2 x
2 x
x e dx = x e − 2 xex dx = x2 ex − 2(xex − ex + C)
= x2 ex − 2xex + 2ex + C1 = ex (x2 − 2x + 2) + C1
5
The Method Of Partial Fractions:
Consider the function f (x) =
3
x
+
4
.
x+4
We can re-express this function by bringing the 2 fractions (i.e., rational
functions) to a common denominator:
4
3(x + 4) + 4(x)
7x + 12
3
+
=
= 2
x x+4
x(x + 4)
x + 4x
R
In its original form, we can find f (x)dx easily:
Z
Z Z
1
4
3
1
+
dx + 4
dx = 3 ln |x| + 4 ln |x + 4| + C
dx = 3
x x+4
x
x+4
f (x) =
(using the substitution u = x + 4 for the second integral, if desired). But
what if this problem had been given to us in the form:
R
Find x7x+12
2 +4x dx.
What would we do then?
The Method of Partial Fractions allows us to reverse the process we used
4
above, in bringing f (x) = x3 + x+4
to the form f (x) = x7x+12
2 +4x . That is, the
(x)
method gives us a way to take a rational function of the form fg(x)
, where
f and g are both polynomial functions, with f having lower degree than g,
and break it up, or “decompose” it into “partial fractions”, i..e, into a sum
of several rational functions, each of which (in the cases we’ll be looking at)
has a constant numerator and a linear denominator, that is, a denominator
of the form ax + b.
Rules for finding the partial fractions decomposition of a rational
function:
1. The numerator polynomial must be of a lower degree than the denominator (i.e. we must have a proper fraction). If it is not, then divide
(perhaps using long division) until the fraction is in proper form.
2. The denominator must have only linear factors, i.e., when factored, the
highest power of x in any factor is 1. (That is, the method we will learn
only works in these cases.)
6
3. A rational function of this type can be broken down into a sum of several
terms, one for each of the factors of the denominator. The form of each
of these terms, i.e. partial fractions, is that the numerator is some
(unknown) constant while the denominator is the corresponding linear
factor of the original denominator. For instance, if the factorization of
the denominator is (x + a)(x + b), then we can decompose the proper
A
B
rational function into x+a
+ x+b
for some values of A and B.
Let’s see how this applies to the problem whose answer we already know.
3
4
That is, let’s see how we decompose x7x+12
2 +4x into x + x+4 using the method of
partial fractions.
We start by factoring the denominator: x2 + 4x = x(x + 4). Then, according
to the Rules above, it is true that
A
B
7x + 12
= +
x(x + 4)
x x+4
for some values of A and B. But by bringing these to a common denominator,
we see that
A
B
A(x + 4) + B(x)
(A + B)x + 4A
+
=
=
x x+4
x(x + 4)
x(x + 4)
so we have
7x + 12
(A + B)x + 4A
=
x(x + 4)
x(x + 4)
and so it must be true that (A + B)x + 4A = 7x + 12.
We see that on the left hand side of this equation we have the constant
term 4A, while on the right hand side the constant term is 12. Thus we must
have 4A = 12, so A = 3. Also, the coefficient of x on the left hand side of
the equation is (A + B), while the coefficient of x on the right hand side is
7, so we must have A + B = 7, and since we have A = 3 then we must have
B = 4. In this way, we see that
7x + 12
3
4
=
+
x2 + 4x
x x+4
Let’s work through a couple of other examples in which we don’t already
know what the answer is.
7
Example 5. Find
R
1
dx
x2 −4
Solution: We first observe that (1) we don’t recognize this integral, (2) substitution doesn’t help and (3) it doesn’t look like integration by parts will help
much either. Since the integrand is a rational function, we try the method
of partial fractions.
1
x2 −4
=
1
(x−2)(x+2)
1
(x−2)(x+2)
=
A
x−2
+
(a) We factor the denominator
into linear factors
B
x+2
(b) Separate the fraction (common
denominator in reverse) and
place new unknowns in the
numerators.
1 = A(x + 2) + B(x − 2)
(c) Multiply both sides by the original
denominator
x = 2 → 1 = A(4) + B(0) → A =
1
4
(d) Make “clever” choices for x in
order to solve for A and B
x = −2 → 1 = A(0) + B(−4) → B = − 41
That is, in the last step above, we can use the “zeroes” of the denominator
function to solve for A and B, since for each of those x-values one of the 2
unknowns is being multiplied by 0.
Having found A and B, we can re-express the original problem as follows:
R 1
R 1
− 41
4
dx =
+ x+2 dx
x2 −4
x−2
=
1
4
R
1
x−2
=
1
4
ln |x − 2| − 41 ln |x + 2| + C
dx + − 41
R
1
dx
x+2
Notice: The substitution rule was used for each integral in the last step.
Also notice: We can check our partial fraction decomposition by bringing the
partial fractions to a common denominator to see that we get the original
function. It is a good idea to always do this.
8
Example 6. Find
R
7x+5
dx.
(x+1)(x−1)(x+2)
Solution: We need to decompose the integrand function into 3 partial fractions, corresponding to the 3 linear factors in the denominator of this rational
function:
7x + 5
A
B
C
=
+
+
(x + 1)(x − 1)(x + 2)
x+1 x−1 x+2
Bringing the right hand side to a common denominator and then equating
numerators, we see that:
7x + 5 = A(x − 1)(x + 2) + B(x + 1)(x + 2) + C(x + 1)(x − 1)
Again, the “zeroes” of the denominator polynomial help us to find the unknowns. When x = −1, we have x + 1 = 0 and so both the B and C terms
on the right hand side are 0 and we get:
7(−1) + 5 = A(−1 − 1)(−1 + 2) ⇒ −2 = A(−2)(1) ⇒ A = 1
Similarly, when x = 1 we have x − 1 = 0 and so both the A and C terms are
0, giving:
12
=2
6
Finally, when x = −2 we have x + 2 = 0 so that both the A term and the B
term are 0 and we get:
7(1) + 5 = B(1 + 1)(1 + 2) ⇒ 12 = B(2)(3) ⇒ B =
7(−2)+5 = C(−2+1)(−2−1) ⇒ −14+5 = C(−1)(−3) ⇒ 3C = −9 ⇒ C = −3
We see that the partial fraction decomposition is:
7x + 5
1
2
3
=
+
−
(x + 1)(x − 1)(x + 2)
x+1 x−1 x+2
(Once again, we can check these calculations by bringing the right hand side
of the decomposition to a common denominator and working out the numerator.)
We can now use this decomposition to evaluate the integral:
R
R 1
7x+5
2
3
dx
=
+
−
dx
(x+1)(x−1)(x+2)
x+1
x−1
x+2
=
R
1
dx
x+1
+2
R
1
dx
x−1
−3
R
1
dx
x+2
= ln |x + 1| + 2 ln |x − 1| − 3 ln |x + 2| + C
9