MATH1110 Lecture tasks  Worked solutions
Integration (III)
Task 3.
Note that the denominator
x
2
+ 2
x
may be factorised as
x
(
x
+ 2), and so using
the method of partial fractions we begin by expressing this fraction in the form
1
x
(
x
+ 2)
≡
A
x
+
B
x
+ 2
=
A
(
x
+ 2) +
Bx
x
(
x
+ 2)
=
(
A
+
B
)
x
+ 2
A
x
(
x
+ 2)
.
This leads to the simultaneous equations
A
+
B
= 0 and 2
A
= 1, from which we can
conclude
A
= 1
/
2 and
B
=

1
/
2. Thus, we obtain
1
x
(
x
+ 2)
=
1
/
2
x

1
/
2
x
+ 2
,
allowing us to evaluate the indefinite integral
Z
1
x
2
+ 2
x
dx
=
Z
1
/
2
x

1
/
2
x
+ 2
dx
=
1
2
ln

x
 
1
2
ln

x
+ 2

+
C.
Task 4.
Note that the denominator
x
2
+ 2
x
+ 1 may be factorised as (
x
+ 1)
2
, and so
using the method of partial fractions we begin by expressing this fraction in the form
1
(
x
+ 1)
2
≡
A
x
+ 1
+
B
(
x
+ 1)
2
=
A
(
x
+ 1) +
B
(
x
+ 1)
2
=
(
Ax
+ (
A
+
B
)
(
x
+ 1)
2
.
This leads to the simultaneous equations
A
= 0 and
A
+
B
= 1, from which we can
conclude
A
= 0 and
B
= 1. Thus, we obtain
1
(
x
+ 1)
2
=
0
x
+ 1
+
1
(
x
+ 1)
2
,
(why did we even bother doing this?)
allowing us to evaluate the indefinite integral
Z
1
x
2
+ 2
x
+ 1
dx
=
Z
1
(
x
+ 1)
2
dx
=

1
x
+ 1
+
C.
(using substitution if needed)
Task 5.
Note that the denominator cannot be factorised over the real numbers, and we
have learned from the last example that partial fractions is a bit pointless when there is
only one factor in the denominator and the numerator is a 1. Instead, we complete the
square on the denominator to rewrite
x
2
+ 2
x
+ 2 as (
x
+ 1)
2
+ 1. Using integration by
substitution (if needed) allows us to evaluate the indefinite integral
Z
1
x
2
+ 2
x
+ 2
dx
=
Z
1
(
x
+ 1)
2
+ 1
dx
= tan

1
(
x
+ 1) +
C.
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