Some Standard Integration Techniques
S. F. Ellermeyer
January 11, 2005
1
The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) tells us that if a function, f ,
is continuous on the interval [a, b] and the function F is any antiderivative
of f on [a, b], then
Z b
f (x) dx = F (b) − F (a) .
a
In order to apply this result, we need to be able to find an antiderivative, F ,
of the integrand, f . Sometimes this is impossible, but there are many cases
in which it can be done.
We will employ the idea of the “indefinite integral” of f , written as
Z
f (x) dx,
to mean the set of all antiderivatives of f on some interval, I. If F is any
particular antiderivative of f on the interval I, then
Z
f (x) dx = F (x) + C (where C can be any real constant).
Since the interval, I, does not appear anywhere in the notation for the indefinite integral, it is not a very good notation. This usually causes no problems,
however, in a course such as this where we mostly consider functions whose
antiderivatives do not depend on which interval in the domain of f we are
considering. For example, there is no ambiguity when we write
Z
1
x2 dx = x3 + C
3
1
or
Z
cos (x) dx = sin (x) + C
because the corresponding differentiation formulas
¶
µ
d 1 3
x + C = x2
dx 3
and
d
(sin (x) + C) = cos (x)
dx
are correct on any subinterval of (−∞, ∞). One notable exception (that
we do encounter very often) is the function f (x) = 1/x. In particular, the
statement
Z
1
dx = ln (x) + C
x
is correct if we are considering the function f (x) = 1/x with domain a
subinterval of (0, ∞), but it is not correct if the domain under consideration
is a subinterval (−∞, 0). In the latter case, a correct statement is
Z
1
dx = ln (−x) + C.
x
We can always be safe (that is, avoid any possible confusion) by using the
formula
Z
1
dx = ln |x| + C.
x
This last statement is correct in both the case that the domain of f (x) = 1/x
is a subinterval (0, ∞) and the case that the domain of f (x) = 1/x is a
subinterval of (−∞, 0).
2
A Basic Beginning Toolbox of Integrals
Below is a list of some basic integrals. These are integrals that should be
memorized. All of the integration techniques that we use to compute more
complicated integrals are aimed at reducing the more complicated integrals
to one of the forms in the basic list.
2
1. If n is any fixed real number (except −1), then
Z
1
xn+1 + C.
xn dx =
n+1
2.
Z
3.
Z
4.
Z
5.
7.
9.
cos (x) dx = sin (x) + C
sin (x) dx = − cos (x) + C
Z
sec2 (x) dx = tan (x) + C
Z
Z
Z
ex dx = ex + C
Z
6.
8.
x−1 dx = ln |x| + C
csc2 (x) dx = − cot (x) + C
sec (x) tan (x) dx = sec (x) + C
csc (x) cot (x) dx = − csc (x) + C
10.
Z
1
√
dx = arcsin (x) + C
1 − x2
11.
Z
1
dx = arctan (x) + C.
1 + x2
3
3
Integration by Substitution
The Chain Rule tells us that is F and g are differentiable functions, then
d
(F (g (x))) = F 0 (g (x)) · g 0 (x) .
dx
Thus,
Z
F 0 (g (x)) · g0 (x) dx = F (g (x)) + C.
If we make the substitutions
u = g (x)
and
du = g0 (x) dx,
then we can write the above integration formula as
Z
F 0 (u) du = F (u) + C
This leads to the idea of “integration by substitution”.
Example 1 Evaluate the indefinite integral
Z
¢8
¡
x2 4x3 + 5 dx.
Solution 2 Make the substitution
u = 4x3 + 5
du = 12x2 dx
and note that
1
du.
12
The above integral can then be written as
Z
1 8
u du.
12
x2 dx =
Since
we see that
Z
Z
1 1 9
1 8
u du =
· u + C,
12
12 9
¢8
¢9
¡
1 ¡ 3
x2 4x3 + 5 dx =
4x + 5 + C.
108
4
Example 3 Evaluate the definite integral
Z π/4
tan2 (x) sec2 (x) dx.
0
Solution 4 Making the substitution
u = tan (x)
du = sec2 (x) dx,
we see that
Z
Z
1
1
2
2
tan (x) sec (x) dx = u2 du = u3 + C = tan3 (x) + C.
3
3
By the FTC, we then have
Z π/4
³π ´ 1
1
1
tan2 (x) sec2 (x) dx = tan3
− tan3 (0) = .
3
4
3
3
0
¡π¢
(Recall that tan 4 = 1 and tan (0) = 0.)
Exercise 5 Evaluate the following indefinite or definite integrals.
1.
Z
2.
¢
¡
sin 3x2 + 2x − 1 · (3x + 1) dx
Z
ln(3)
√
ex ex + 4 dx
0
3.
Z
4.
3x2 + 8x − 6
dx
x3 + 4x2 − 6x − 2
Z
π/4
tan (x) dx
0
5.
Hint: e2x = (ex )2 .
Z
ex
dx
e2x + 1
5
4
Integration By Parts
The Product Rule tells us that if f and g are differentiable functions, then
d
(f (x) · g (x)) = f (x) · g 0 (x) + f 0 (x) · g (x) .
dx
Thus
or
Z
Z
(f (x) · g 0 (x) + f 0 (x) · g (x)) dx = f (x) · g (x) + C
0
f (x) · g (x) dx +
Making the substitutions
Z
u
du
v
dv
f 0 (x) · g (x) dx = f (x) · g (x) + C
=
=
=
=
f (x)
f 0 (x) dx
g (x)
g 0 (x) dx,
we can write the above integration formula as
Z
Z
u dv + v du = uv + C
or as
Z
u dv = uv −
Z
v du + C.
(It is not really necessary to write the “+C” at the end of the above formula,
since the indefinite integral on the right automatically includes a “+C”.)
The above formula is what we call
R the formula for integration by parts. It
is
R used in cases where the integral v du is easy to compute, but the
R integral
u dv is not. Specifically,Rit is used when we want to compute u dv, and
we do this via computing v du.
Example 6 Evaluate the indefinite integral
Z
x cos (x) dx.
6
Solution 7 The above integral has the form
R
u dv where
u=x
dv = cos (x) dx
.
du = dx v = sin (x)
Using the integration by parts formula, we obtain
Z
Z
x cos (x) dx =
u dv
Z
= uv − v du
Z
= x sin (x) − sin (x) dx
= x sin (x) + cos (x) + C.
We conclude that
Z
x cos (x) dx = x sin (x) + cos (x) + C.
Sometimes, it is necessary to use integration by parts more than once, as
in the next example.
Example 8 Evaluate the indefinite integral
Z
ex cos (x) dx.
Solution 9 Make the following substitutions
u = ex
dv = cos (x) dx
.
x
du = e dx v = sin (x)
Then
Z
x
e cos (x) dx =
Z
u dv
Z
= uv − v du
Z
x
= e sin (x) − ex sin (x) dx.
7
We now see that
Z
x
x
e cos (x) dx = e sin (x) −
Z
ex sin (x) dx.
(1)
At this point,
R it appears that integration by parts has perhaps beenR of no
help, because v du appears to be no easier than the original problem, u dv.
However, we continue (using integration by parts again) with the new problem
Z
ex sin (x) dx.
Here we make the substitutions
u = ex
dv = sin (x) dx
x
du = e dx v = − cos (x)
and obtain
Z
x
e sin (x) dx = uv −
Z
v du
x
= −e cos (x) +
Thus,
Z
x
x
e sin (x) dx = −e cos (x) +
Z
Z
ex cos (x) dx.
ex cos (x) dx.
(2)
Now, combining the above results (equations (1) and (2)), we observe that
Z
Z
x
x
e cos (x) dx = e sin (x) − ex sin (x) dx
µ
¶
Z
x
x
x
= e sin (x) − −e cos (x) + e cos (x) dx
Z
x
x
= e sin (x) + e cos (x) − ex cos (x) dx
or, in summary,
Z
Z
x
x
x
e cos (x) dx = e sin (x) + e cos (x) − ex cos (x) dx.
8
(3)
Adding
R
ex cos (x) dx to both sides of equation (3) gives us
Z
2 ex cos (x) dx = ex sin (x) + ex cos (x) .
We then divide both sides of the above equation by 2 to obtain
Z
1
ex cos (x) dx = ex (sin (x) + cos (x)) + C.
2
Exercise 10 Evaluate the following integrals.
1.
Z
x ln (x) dx
Z
x sin (x) dx
Z
ex sin (x) dx
Hint: Let u = ln (x) and dv = x dx.
2.
3.
4.
Z
1
xe2x dx
0
5.
6.
Z
x sin (4x + 2) dx
Z
π/2
x2 cos (3x) dx
0
7.
Z
ln (x) dx
9
8.
Z
(ln (x))2 dx
Hint: Use your result from exercise 7.
9.
Z
10.
arctan (x) dx
Z
x arctan (x) dx
Z
e3x cos (4x) dx
Hint: Use your result from exercise 9 and you may also find it useful
to use the fact that
1
x2
=
1
−
.
1 + x2
1 + x2
11.
5
Trigonometric Integrals
Let’s consider how to evaluate integrals of the form
Z
sinm (x) cosn (x) dx
where m and n are positive integers. The right approach to take depends on
whether m and n are even or odd integers.
Example 11 Evaluate
Z
sin5 (x) cos4 (x) dx.
Solution 12 The key to evaluating this integral is to notice that we have an
odd power of sine in the integrand. We would like to “reserve” a factor of
10
sin (x) so that we can make the substitution u = cos (x), du = − sin (x) dx.
We can do this by rewriting the integrand as
sin5 (x) cos4 (x) = cos4 (x) sin4 (x) sin (x)
¡
¢2
= cos4 (x) sin2 (x) sin (x)
¡
¢2
= cos4 (x) 1 − cos2 (x) sin (x)
¡
¢
= cos4 (x) 1 − 2 cos2 (x) + cos4 (x) sin (x)
¢
¡
= cos4 (x) − 2 cos6 (x) + cos8 (x) sin (x) .
Now using the substitution u = cos (x), du = − sin (x) dx, we have
Z
Z
¡ 4
¢
5
4
sin (x) cos (x) dx =
cos (x) − 2 cos6 (x) + cos8 (x) sin (x) dx
Z
¡ 4
¢
= −
u − 2u6 + u8 du
Z
¡ 4
¢
=
−u + 2u6 − u8 du
1
2
1
= − u5 + u7 − u9 + C
5
7
9
1
1
2
5
= − cos (x) + cos7 (x) − cos9 (x) + C.
5
7
9
The general rule to be learned from the above example is that if either
m or n is an odd integer in the integral
Z
sinm (x) cosn (x) dx,
then “reserve” one factor of sine or cosine (whichever has the odd power)
and use the Pythagorean identity sin2 (x) + cos2 (x) = 1 to rewrite the rest
of the integrand in terms of either sine or cosine (whichever has not been
“reserved”). If both m and n are odd, then we can choose to reserve either
a factor of sine or a factor of cosine.
If both m and n are even, then we must make use of the trigonometric
identities
1 − cos (2x)
2
1
+
cos
(2x)
cos2 (x) =
.
2
sin2 (x) =
11
The identity
sin (2x) = 2 sin (x) cos (x)
might also come in handy.
This idea is illustrated in the following example.
Example 13 Evaluate
Z
sin4 (x) cos2 (x) dx.
Solution 14 We rewrite the integrand as follows:
sin4 (x) cos2 (x) = (sin (x) cos (x))2 sin2 (x)
¶2 µ
¶
µ
1 − cos (2x)
1
sin (2x)
=
2
2
µ
¶
1 1
1 2
sin (2x)
− cos (2x)
=
4
2 2
1
1 2
sin (2x) − sin2 (2x) cos (2x) .
=
8
8
At this point, note that the integral of
1 2
sin (2x) cos (2x)
8
will not be hard to do because it has an odd power of cosine. We will leave
this term as it is, but we still need to put the term
1 2
sin (2x)
8
into a form that we can integrate. Continuing, we have
1
1 2
sin (2x) − sin2 (2x) cos (2x)
8µ
8 ¶
1
1 1 − cos (4x)
=
− sin2 (2x) cos (2x)
8
2
8
µ
¶
1
1 1 1
=
− cos (4x) − sin2 (2x) cos (2x)
8 2 2
8
1
1
1
cos (4x) − sin2 (2x) cos (2x) .
=
−
16 16
8
sin4 (x) cos2 (x) =
12
We can now compute the desired integral by breaking it into three integrals:
¶
Z µ
Z
1 2
1
1
4
2
cos (4x) − sin (2x) cos (2x) dx
sin (x) cos (x) dx =
−
16 16
8
Z
Z
1
1
dx −
cos (4x) dx
=
16
16
Z
1
sin2 (2x) cos (2x) dx
−
8
Z
1
1
1
=
x−
sin (4x) −
sin2 (2x) cos (2x) dx.
16
64
8
To do the last integral (which contains an odd power of cosine), we use the
substitution u = sin (2x), du = 2 cos (2x) dx. This gives us
Z
Z
1
2
sin (2x) cos (2x) dx =
u2 du
2
1 3
=
u +C
6
1 3
=
sin (2x) + C.
6
Our final result is
Z
1
1
1
sin4 (x) cos2 (x) dx = x −
sin (4x) −
sin3 (2x) + C.
16
64
48
We will now check our result graphically: Below are graphs (drawn using
Maple) of the functions
F (x) =
1
1
1
x−
sin (4x) −
sin3 (2x)
16
64
48
and
f (x) = sin4 (x) cos2 (x) .
By comparing these graphs, it looks reasonable that F 0 (x) = f (x).
13
Exercise 15 Evaluate the following integrals:
1.
Z
2.
3.
Z
Z
4.
5.
cos4 (x) sin (x) dx
sin3 (x) cos3 (x) dx
Z
Z
cos4 (x) dx
sin5 (x) dx
sin2 (x) cos2 (x) dx
14
6
Trigonometric Substitution
√
a2 − x2 ,
In
dealing
with
integrals
whose
integrands
contain
terms
of
the
form
√
√
2
2
2
2
a + x , or x − a (where a is a constant), it is often helpful to make a
trigonometric substitution.
√
If we have the term a2 − x2 , then the substitution to try is
x = a sin (θ) .
In this case we have
dx = a cos (θ) dθ
and also
x2 = a2 sin2 (θ)
so
q
√
a2 − x2 =
a2 − a2 sin2 (θ)
q ¡
¢
=
a2 1 − sin2 (θ)
p
= a cos2 (θ)
= a cos (θ) .
(This removes the square root from the problem.) This is illustrated in the
following example.
Example 16 Use integration to show that the area of a circle of radius r is
πr2 .
Solution 17 The equation for the circle of radius r centered at the√origin is
x2 + y 2 = r2 . The equation for the upper half of this circle is y = r2 − x2 .
Thus, the area of the circle is
Z r√
r2 − x2 dx.
A=2
−r
To evaluate this integral, we make the trigonometric substitution
x = r sin (θ) .
This gives us
dx = r cos (θ) dθ
15
and also gives us
Now we have
q
√
2
2
r − x = r2 − r2 sin2 (θ) = r cos (θ) .
A = 2
Z
√
r2 − x2 dx
r
−r
π/2
= 2
Z
= 2r
r cos (θ) r cos (θ) dθ
−π/2
Z π/2
2
cos2 (θ) dθ
−π/2
π/2
1 + cos (2θ)
dθ
2
−π/2
¶
Z π/2 µ
1 1
2
2r
+ cos (2θ) dθ
2 2
−π/2
µ
¶¯θ=π/2
¯
1
1
2
θ + sin (2θ) ¯¯
2r
2
4
θ=−π/2
´ ³ π
´´
³³ π
+0 − − +0
2r2
4
³ π 4´
2r2
2
πr2 .
= 2r2
=
=
=
=
=
Z
The relevant trigonometric
√ substitution for dealing with integrands that
2
contain terms of the form a2 +
√x is x = a tan (θ) and the relevant one for
dealing with terms of the form x2 − a2 is x = a sec (θ).
Example 18 Evaluate the integral
Z
1
√
dx.
4 + x2
Solution 19 We make the trigonometric substitution
x = 2 tan (θ) .
16
This gives us
dx = 2 sec2 (θ) dθ
and
q
√
2
4+x =
4 + 4 tan2 (θ)
q
=
4 (1 + tan2 (θ))
p
= 2 sec2 (θ)
= 2 sec (θ) .
We now obtain
Z
Z
Z
1
1
2
√
dx =
· 2 sec (θ) dθ = sec (θ) dθ.
2 sec (θ)
4 + x2
The integral
Z
sec (θ) dθ
is very tricky to do (if you don’t know the trick) and perhaps this trick is one
that you should remember! Here’s how to do it: We observe that
sec (θ) =
Thus
Z
sec (θ) sec (θ) + tan (θ)
sec2 (θ) + sec (θ) tan (θ)
·
=
.
1
sec (θ) + tan (θ)
sec (θ) + tan (θ)
sec (θ) dθ =
Z
sec2 (θ) + sec (θ) tan (θ)
dθ.
sec (θ) + tan (θ)
We now make the simple substitution
u = sec (θ) + tan (θ) .
Then
and we have
Z
¢
¡
du = sec2 (θ) + sec (θ) tan (θ) dθ
sec2 (θ) + sec (θ) tan (θ)
dθ =
sec (θ) + tan (θ)
17
Z
1
du = ln |u| + C.
u
To summarize from the beginning, we have
Z
Z
Z
1
1
√
du = ln |u| + C.
dx = sec (θ) dθ =
u
4 + x2
Our answer is in terms of u, but we would like it to be in terms of x, so we
need to substitute back:
Z
1
√
dx = ln |u| + C
4 + x2
= ln |sec (θ) + tan (θ)| + C
¯q
¯
¯
¯
2
¯
= ln ¯ 1 + tan (θ) + tan (θ)¯¯ + C
¯s
¯
¯
¯
µ ¶2
¯
¯
1
1
= ln ¯¯ 1 +
x + x¯¯ + C.
2
2 ¯
¯
In conclusion,
Z
¯
¯r
¯
1
1 2 1 ¯¯
¯
√
dx = ln ¯ 1 + x + x¯ + C.
¯
4
2 ¯
4 + x2
We can make this answer look nicer: Note that
r
r
4 + x2
1√
1 2
=
1+ x =
4 + x2
4
4
2
so
r
´
1√
1 ³√
1
1
1
1 + x2 + x =
4 + x2 + x =
4 + x2 + x
4
2
2
2
2
and
¯
¯r
¯ ³
¯ ¯
¯√
¯
¯
´¯¯
¯1 √
¯1¯
1 2 1 ¯¯
¯
¯
¯
2
2
¯
¯
¯
¯
4 + x + x ¯ = ln ¯ ¯ + ln ¯ 4 + x + x¯ .
ln ¯ 1 + x + x¯ = ln ¯
¯
4
2 ¯
2
2
The
√ ln |1/2| can be “absorbed” in the integration constant C. Also, since
4 + x2 + x is always positive, we can get rid of the absolute value. This
gives us the result
Z
³√
´
1
√
dx = ln
4 + x2 + x + C.
4 + x2
18
Let us check that this result is correct by differentiation:
µ
¶
´´
¢
1¡
1
d ³ ³√
2 −1/2
2
ln
4+x
4+x +x
= √
· 2x + 1
·
dx
2
4 + x2 + x
µ
¶
x
1
+1
= √
· √
4 + x2 + x
4 + x2
Ã
!
√
1
x + 4 + x2
√
= √
·
4 + x2 + x
4 + x2
1
= √
.
4 + x2
Exercise 20 Evaluate the following integrals:
7
1.
Z
1
√
dx
9 − x2
2.
Z
1
√
dx
9 + x2
3.
Z √
25 − x2 dx
4.
Z √
4 + x2 dx
5.
Z
1
√
dx
x2 − 9
Partial Fraction Decomposition
To evaluate integrals of the form
Z
p(x)
dx
q(x)
19
where p and q are polynomials with the degree of p less than the degree of
q, we sometimes first need to write p (x) /q (x) as a sum of rational functions
whose denominators are powers of linear and irreducible quadratic factors of
q. This is called doing a “partial fraction decomposition” of the integrand.
We illustrate this technique by looking at some examples.
Example 21 Consider the problem of computing the indefinite integral
Z
x+1
dx.
(4)
3
x + 2x2 − 3x
In factored form the integrand is
x+1
.
x(x − 1)(x + 3)
We attempt to decompose this rational function as
x+1
A
B
C
= +
+
x(x − 1)(x + 3)
x x−1 x+3
(5)
where A, B, and C are constants to be determined. To determine the appropriate constants, we multiply both sides of equation (5) by the denominator
of the left hand side to obtain
x + 1 = A(x − 1)(x + 3) + Bx(x + 3) + Cx(x − 1).
(6)
Substitution of x = 0 into (6) gives
0 + 1 = A(0 − 1)(0 + 3) + B(0)(0 + 3) + C(0)(0 − 1)
from which we conclude that A = −1/3.
Substitution of x = 1 into (6) gives
1 + 1 = A(1 − 1)(1 + 3) + B(1)(1 + 3) + C(1)(1 − 1)
which yields that B = 1/2.
Similarly, substitution of x = −3 into (6) gives C = −1/6.
Substituting the values of A, B, and C which have been found back into
(5), we obtain the partial fraction decomposition
−1/3
1/2
−1/6
x+1
=
+
+
,
x(x − 1)(x + 3)
x
x−1 x+3
20
(7)
which can also be written as
x+1
1
=−
x(x − 1)(x + 3)
6
µ
¶
3
2
1
−
+
.
x x−1 x+3
The indefinite integral (4) can now be computed:
¶
Z µ
Z
2
1
1
x+1
3
+
dx
dx = −
−
x3 + 2x2 − 3x
6
x x−1 x+3
1
= − (2 ln |x| − 3 ln |x − 1| + ln |x + 3|) + C
6
¢
1¡
= − ln x2 − ln |x − 1|3 + ln |x + 3| + C
6 µ
¶
1
x2 |x + 3|
+C
= − ln
6
|x − 1|3
Example 22 Let us write the rational expression
1
x(x + 1)(x + 2)3
in partial fraction form.
Like the rational expression which was the subject of Example 21, the
rational expression
1
x(x + 1)(x + 2)3
has a denominator which is a product of linear factors. Since the factor x + 2
has multiplicity three, the partial fraction decomposition will have the form
1
A
E
B
C
D
.
+
+
+
3 =
2 +
x x + 1 x + 2 (x + 2)
x(x + 1)(x + 2)
(x + 2)3
(8)
Multiplication of both sides of (8) by the denominator of the left hand side
yields
1 = A(x + 1)(x + 2)3 + Bx(x + 2)3 + Cx(x + 1)(x + 2)2
+Dx(x + 1)(x + 2) + Ex(x + 1).
When x = 0 is substituted into (9), we obtain
1 = A(0 + 1)(0 + 2)3 ,
21
(9)
which implies that A = 1/8.
Substitution of x = −1 and x = −2 yield, respectively, that B = −1 and
E = 1/2
Hence, equation (9) becomes
1 =
1
(x + 1)(x + 2)3 − x(x + 2)3 + Cx(x + 1)(x + 2)2
8
1
+Dx(x + 1)(x + 2) + x(x + 1).
2
(10)
When x = −3 is substituted into (10), we obtain
1 =
1
(−3 + 1)(−3 + 2)3 − (−3)(−3 + 2)3 + C(−3)(−3 + 1)(−3 + 2)2
8
1
+ D(−3)(−3 + 1)(−3 + 2) + (−3)(−3 + 1)
2
which simplifies to
1
C −D = .
8
Substitution of x = 1 into (10) yields, after simplification,
3C + D =
27
.
8
(11)
(12)
Solving equations (11) and (12) simultaneously yields that C = 7/8 and
D = 3/4.
Substitution of the values of A, B, C, D, and E which were found into
(8) yields the partial fraction decomposition
¶
µ
1
8
7
6
1 1
4
. (13)
+
+
=
+
−
8 x x + 1 x + 2 (x + 2)2 (x + 2)3
x(x + 1)(x + 2)3
Example 23 As our final example, let us write the rational expression
x4
x3 − 1
in partial fraction form.
22
Since the degree of the polynomial in the numerator is not less than the
degree of the polynomial in the denominator, we must first perform long division before attempting a partial fraction decomposition. After performing
the long division, we obtain
x4
x
=x+ 3
.
3
x −1
x −1
(14)
The second term on the right hand side of (14) can be written in factored
form as
x
x
.
=
(15)
3
x −1
(x − 1)(x2 + x + 1)
Since the factor x2 +x+1 which appears in equation (15) is an irreducible
quadratic factor, we attempt to decompose (15) as a sum of the form
x
A
Bx + C
=
+ 2
.
2
(x − 1)(x + x + 1)
x−1 x +x+1
(16)
After multiplication of both sides by the denominator of the left hand side,
equation (16) becomes
x = A(x2 + x + 1) + (Bx + C)(x − 1).
(17)
Substitutions of x = 1, x = 0, and x = −1 into (17) yield, respectively,
that A = 1/3, C = 1/3, and B = −1/3.
Going back to (14), we obtain the partial fraction decomposition
µ
¶
1
−x + 1
x4
1
+
.
=x+
x3 − 1
3 x − 1 x2 + x + 1
Exercise 24
1. Obtain partial fraction decompositions of the following
rational expressions. Then evaluate the indefinite integral of each expression (if you can).
(a)
x−2
x2 + 3x + 2
(b)
x2
23
x−2
+ 4x + 4
(c)
x(x2
2
− x + 2)
(d)
2
(x2 − 2x + 1)(x2 + 3x + 2)2
2. Use the results in Examples 22 and 23 to evaluate the indefinite integrals
Z
1
dx
x(x + 1)(x + 2)3
and
Z
x4
dx.
x3 − 1
(You might find the second one to be challenging!)
24