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MTH 122 — Calculus II
Essex County College — Division of Mathematics and Physics
1
Lecture Notes #5 — Sakai Web Project Material
1
Tough Problems?
1. Supposed you’re asked to calculate the integral
Z
1
x
2
+ 4
x
+ 13
d
x
?
Many would just stop because they can’t factor. Look over the prior notes and you may
realize that this quadratic factor in the numerator is irreducible. So it is probably related
to the only irreducible form you know:
Z
1
x
2
+ 1
d
x.
Let’s proceed to get it in that form.
Z
1
x
2
+ 4
x
+ 13
d
x
=
Z
1
x
2
+ 4
x
+ 4 + 9
d
x
=
Z
1
(
x
+ 2)
2
+ 9
d
x
=
1
9
Z
1
((
x
+ 2)
/
3)
2
+ 1
d
x
Now let
u
= (
x
+ 2)
/
3, then 3 d
u
= d
x
.
1
9
Z
1
((
x
+ 2)
/
3)
2
+ 1
d
x
=
1
3
Z
*
1
u
2
+ 1
d
u
=
1
3
arctan
u
+
C
=
1
3
arctan
x
+ 2
3
+
C
2. I’ve noticed that even a slight variation in the way a question is worded can prevent
even the best students from following through. Let’s say that you’re asked to ﬁnd the
antiderivative of
x
2

x
+ 2
x
3

x
2
+
x

1
that passes through the point (2
,
6). What would you do?
Well, here’s what I’d do.
Z
x
2

x
+ 2
x
3

x
2
+
x

1
d
x
=
Z
x
2

x
+ 2
(
x

1) (
x
2
+ 1)
d
x
1
This document was prepared by Ron Bannon (
ron.bannon@mathography.org
) using L
A
T
E
X2
ε
. Last revised
September 8, 2009.
1
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View Full DocumentCertainly a partial fraction decomposition of the form:
x
2

x
+ 2
(
x

1) (
x
2
+ 1)
=
A
x

1
+
Bx
+
C
x
2
+ 1
.
Using the method discussed in class, we have:
x
2

x
+ 2 =
A
(
x
2
+ 1
)
+ (
Bx
+
C
) (
x

1)
.
Let
x
= 1 and we have:
2 = 2
A
⇒
A
= 1
.
Now let
x
= 0 and we have:
2 = 1

C
⇒
C
=

1
.
Now let
x
= 2 and we have:
4 = 5 + (2
B

1)
⇒
B
= 0
.
So now we can continue the integration.
2
Z
x
2

x
+ 2
x
3

x
2
+
x

1
d
x
=
Z
x
2

x
+ 2
(
x

1) (
x
2
+ 1)
d
x
=
Z
1
x

1

1
x
2
+ 1
d
x
= ln

x

1
 
arctan
x
+
k
The general antiderivative is
f
(
x
) = ln

x

1
 
arctan
x
+
k,
and to ﬁnd the constant
k
you’ll need to use the point (2
,
6),
i.e.
f
(2) = 6.
6 = ln

1
 
arctan 2 +
k
6 + arctan 2 =
k
Finally we have:
f
(
x
) = ln

x

1
 
arctan
x
+ 6 + arctan2
3. Okay, enough already! Not really, we could actually go on for days if not years. Very last
one, today at least. Integrate!
Z
e
x
e
2
x
+
e
x

2
d
x
First let
u
=
e
x
, then d
u
=
e
x
d
x
.
Z
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 Calculus, Division

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