INTEGRATION BY PARTIAL FRACTIONS AND STRATEGY FOR
INTEGRATION
Z
Last time we left off explaining how to compute the integral
(x4 + x2 + 1)dx
x(1 + x2 )2
using partial fractions. The bottom line is that we can write
x4 + x2 + 1
Dx + E
A Bx + C
+
= +
.
2
2
2
x(x + 1)
x (1 + x ) (1 + x2 )2
Example: Let us now consider more difficult cases. Let us consider the integral
Z
dx
, for which we have the partial fraction expansion
(x + 1)(1 + x2 )
1
A B1 x + C1
Bk x + Ck
= +
+ ... +
,
2
k
2
(x + 1)(1 + x )
x
1+x
(1 + x2 )k
and note that for any j = 1, . . . , k
1
Z
xdx
ln(1 + x2 )
for j = 1.
2
=
1−j
2 1−j
2
j
(1
+
x
)
for j > 1
(1 + x )
2
We can also compute
Z
dx
, for j = 1, . . . , k,
(1 + x2 )j
by means of the substitution u = tan(θ), in fact
Z
Z
Z
dx
sec2 (θ)dθ
=
= cos2(j−1) (θ)dθ.
2
j
2j
(1 + x )
sec (θ)
Example: Let us now compute the integral
Z
dx
.
2
(x − 2) (x2 + 2x + 5)2
We start by writing the partial fraction expansion of
1
(x − 2)2 (x2 + 2x + 5)2
1
(x −
2)2 (x2
+ 2x + 5)
Ax + B
Dx + C
E
F
= 2
+ 2
+
+
.
2
(x + 2x + 5) (x + 2x + 5)
(x − 2) (x − 2)2
Before solving for the coefficients A, . . . , F , we must make sure that we know how to
integrate all functions appearing in the above expansion. It may be a good idea to
complete a square in the denominator x2 + 2x + 5, i.e.,
x2 + 2x + 5 = (x + 1)2 + 4,
1
2
INTEGRATION BY PARTIAL FRACTIONS AND STRATEGY FOR INTEGRATION
so that we can use the substitution u = x + 1 and then
Z
Z
xdx
x dx
=
x2 + 2x + 5
(x + 1)2 + 4
Z
Z
(u + 1)du
udu
du
=
=
+ 2
2
2
u +4
(u + 4)
+ 4}
|u {z
u=2 tan(θ)
1
1
ln(u2 + 4) + arctan
2
2
2 1
1
x+1
2
= ln(x + 2x + 5) + arctan
+ C.
2
2
2
=
u
We also have
Z
du
1
x+1
dx
=
= arctan
+ C.
(x2 + 2x + 5)
(u2 + 4)
2
2
For the rest of the terms we find that
Z
Z
Z
udu
du
xdx
=
+
2
2
2
2
2
(x + 2x + 5)
(u + 4)
(u + 4)2
Z
1
du
= − (u2 + 4)−1 +
,
2
2
(u + 4)2
Z
and with the substitution u = 2 tan(θ) we have
Z
Z
Z
1
1
du
sec2 (θ)dθ
=
=
cos2 (θ)dθ
(u2 + 4)2
2
sec4 (θ)
2
Z
1
(1 + cos(2θ)) dθ
=
4
θ 1
= + sin(2θ) + C
4 8
u 1
1
= arctan
+ sin(2θ) + C
4
4
2
1
x+1
1
= arctan
+ sin(θ) cos(θ) + C
4
2
2
1
x+1
1
2
u
√
√
= arctan
+
+C
4
2
2
u2 + 4
4 + u2
1
x+1
u
= arctan
+ 2
+C
4
2
(u + 4)
1
x+1
x+1
+ 2
= arctan
+C
4
2
x + 2x + 5
1
x+1
x+1
= arctan
+ 2
+C
4
2
x + 2x + 5
INTEGRATION BY PARTIAL FRACTIONS AND STRATEGY FOR INTEGRATION
3
It follows that
Z
Z
Z
xdx
udu
du
=
+
(x2 + 2x + 5)2
(u2 + 4)2
(u2 + 4)2
1
x+1
(x + 1)
x+1
= arctan
−
+ 2
+ C.
2
4
2
2(x + 2x + 5) (x + 2x + 5)
x+1
(x + 1)
1
+
+ C.
= arctan
2
4
2
2(x + 2x + 5)
We have also shown that
Z
dx
x+1
x+1
1
+ 2
+ C.
= arctan
2
2
(x + 2x + 5)
4
2
x + 2x + 5
Remark: Consider the function
by
1
,
(x+1)(x+2)(x−1)3
its partial fraction expansion is given
1
A
B
C
D
E
=
+
+
+
+
,
3
2
(x + 1)(x + 2)(x − 1)
(x + 1) (x + 2) (x − 1) (x − 1)
(x − 1)3
and note that
1
A
B
3
= (x − 1)
+
+C(x − 1)2 + D(x − 1) + E
(x + 1)(x + 2)
(x + 1) (x + 2)
{z
}
|
f (x)
0
and observe that f (x) has a triple root at (x − 1) and therefore f (1) = f (1) =
00
f (1) = 0. This observation is what allows us to compute the coefficients C, D, E.
Strategy for Integration
(1) Simplify. Examples Another example is given by
Z
Z
tan5 (x)dx
= tan5 (x) sec2 (x)dx.
cos2 (x)
(2) Use u-substitutions whenever possible. For example,
Z
dx
,
x (ln(x))5
Use for example u = ln(x). Another example is given by
Z
x2 dx
,
1 + x6
try u = x3 so that
Z
Z
x2 dx
1
du
1
=
= arctan(x3 ) + C.
6
2
1+x
3
1+u
3
(3) Classify into the following types
4
INTEGRATION BY PARTIAL FRACTIONS AND STRATEGY FOR INTEGRATION
Z
m
Z
sin (x) cos (x)dx,
tann (x) secm (x)dx. Another exZ
ample is given by
sin(2x) cos(4x)dx in which the integrand can be
• Trigonometric
n
simplified by using
sin(nx) cos(mx) =
1
(sin((n + m)x) + sin((n − m)x))
2
and then
1
sin(2x) = cos(4x) = (sin(6x) − sin(2x)) .
2
√
• Integrals involving ±a2 ± x2 use trigonometric substitutions.
• Rational functions: use partial fractions. For example, compute
Z
dx
,
3
(x − 1)2
Recall the identity
(x3 − 1)2
,
x−1
which comes from adding the first three terms of the geometric progression 1, x, x2 , . . . xn , . . .. Let us now compute
Z
Z
dx
dx
=
dx
3
2
2
(x − 1)
(x − 1) (x2 + x + 1)2
1 + x + x2 =
and we can use partial fractions by writing
A
B
Cx + D
Ex + F
1
=
+
+
+
.
(x − 1)2 (x2 + x + 1)2
(x − 1) (x − 1)2 (1 + x + x2 ) (1 + x + x2 )2
√
√
• Integrals involving n ax + b : Try the substitution u = n ax + b. For
example
Z √
3
1 + xdx
,
x
√
If u = 3 1 + x then x = u3 − 1 and dx = 3u2 du and then
Z √
Z
Z 3
1 + xdx
u3 du
1
=3
=3
1+ 3
du.
x
u3 − 1
u −1
Remark: The resulting integrals are always integrals of rational functions.
• Integration by parts
• Relate to integrals previously studied. For example
Z
Z
d
sin(x)
sin(x) cos(x)e
dx = sin(x)esin(x) sin(x)dx,
dx
INTEGRATION BY PARTIAL FRACTIONS AND STRATEGY FOR INTEGRATION
and using u = sin(x) we obtain
Z
Z
sin(x)
sin(x) cos(x)e
dx = ueu du,
which we can solve using integration by parts.
5