MAC 2312 Quiz 3
January 31, 2012
SOLUTIONS
1.
Evaluate the indefinite integral
integraldisplay
x
4
x
2

9
dx.
Solution:
Begin by using polynomial long division to writhe the integrand as follows:
x
4
x
2

9
=
x
2
+ 9 +
81
x
2

9
.
The given integral is therefore
integraldisplay
x
4
x
2

9
dx
=
integraldisplay
x
2
dx
+ 9
integraldisplay
dx
+ 81
integraldisplay
1
x
2

9
dx
=
x
3
3
+ 9
x
+ 81
integraldisplay
1
x
2

9
dx
+
C.
(1)
It remains to compute
integraltext
1
/
(
x
2

9)
dx
. To do this, find the partial fraction decom
position of 1
/
(
x
2

9) as follows. First, write
1
x
2

9
=
1
(
x

3)(
x
+ 3)
=
A
x
+ 3
+
B
x

3
,
(2)
where
A
and
B
are to be determined. To determine
A
and
B
, multiply both sides of
equation
(2) by (
x
+ 3)(
x

3) then rearrange terms to get
1
=
A
(
x

3) +
B
(
x
+ 3)
=
(
A
+
B
)
x
+ 3(
B

A
)
.
From this, by matching coefficients of likepowers of
x
, we conclude that
A
+
B
= 0
and
3(
B

A
) = 1
.
Solving for
A
and
B
yields
A
=

1
/
6 and
B
= 1
/
6. Use these values of
A
and
B
in
equation
(2) then integrate to get
integraldisplay
1
x
2

9
dx
=

1
6
integraldisplay
1
x
+ 3
dx
+
1
6
integraldisplay
1
x

3
dx
=

1
6
ln

x
+ 3

+
1
6
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 Spring '08
 Bonner
 Calculus, Division, dx

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