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Create assignment, 57611, Homework 14, Mar 23 at 9:27 pm
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version 888
CalC8e58s
54:06, calculus3, multiple choice,
>
1 min,
wordingvariable.
001
Determine the indefnite integral
I
=
Z
2
x
ln (2 +
x
)
dx.
1.
I
= (
x
2

4) ln (2 +
x
) + 2
x
+
1
2
x
2
+
C
2.
I
=
1
2
(
x
2

4) ln (2 +
x
)

x

1
4
x
2
+
C
3.
I
=
1
2
(
x
2

4) ln (2 +
x
) +
x

1
4
x
2
+
C
4.
I
= 2(
x
2

4) ln (2 +
x
)

4
x
+
x
2
+
C
5.
I
= (
x
2

4) ln (2 +
x
) + 2
x

1
2
x
2
+
C
correct
6.
I
= 2(
x
2

4) ln (2 +
x
) + 4
x

x
2
+
C
Explanation:
Integrate frst by parts. Then
I
=
x
2
ln (2 +
x
)

Z
x
2
2 +
x
dx.
To evaluate the integral on the right we divide:
x
2
2 +
x
=
2
x
+
x
2

2
x
2 +
x
=
x

2
x
2 +
x
=
x

2 +
4
2 +
x
.
In this case,
Z
x
2
2 +
x
dx
=
Z
n
x

2 +
4
2 +
x
o
dx
=
1
2
x
2

2
x
+ 4 ln (2 +
x
)
.
Consequently,
I
= (
x
2

4) ln (2 +
x
) + 2
x

1
2
x
2
+
C
with
C
an arbitrary constant.
keywords: indefnite integral, integration by
parts, natural log, polynomial long division
CalC8a08d
54:02, calculus3, multiple choice,
>
1 min,
wordingvariable.
002
±ind the indefnite integral
Z
3
x
(ln
x
)
2
dx.
1.
3
2
x
2
‡
(ln
x
)
2
+ ln
x
+
1
2
·
+
C
2.
6
x
2
‡
(ln
x
)
2

ln
x
+
1
2
·
+
C
3.
3
2
x
2
‡
(ln
x
)
2

ln
x
+
1
2
·
+
C
correct
4.
3
x
2
‡
(ln
x
)
2

ln
x
+
1
2
·
+
C
5.
3
2
x
2
‡
(ln
x
)
2

ln
x

1
2
·
+
C
Explanation:
AFter integration by parts,
Z
x
(ln
x
)
2
dx
=
1
2
x
2
(ln
x
)
2

Z
x
2
1
x
ln
xdx
=
1
2
x
2
(ln
x
)
2

Z
x
ln
xdx.
But aFter integration by parts once again,
Z
x
ln
xdx
=
1
2
x
2
ln
x

1
2
Z
x
2
1
x
dx
=
1
2
x
2
ln
x

1
2
Z
xdx
=
1
2
x
2
ln
x

1
4
x
2
+
C.
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View Full Document Create assignment, 57611, Homework 14, Mar 23 at 9:27 pm
2
Thus
Z
x
(ln
x
)
2
dx
=
1
2
x
2
(ln
x
)
2

1
2
x
2
ln
x
+
1
4
x
2
+
C.
Consequently,
Z
3
x
(ln
x
)
2
dx
=
3
2
x
2
‡
(ln
x
)
2

ln
x
+
1
2
·
+
C
.
keywords: integration by parts, log function
CalC8b22exam
54:03, calculus3, multiple choice,
>
1 min,
wordingvariable.
003
Evaluate the deFnite integral
I
=
Z
π/
4
0
4 tan
4
xdx.
1.
I
=
π

8
3
correct
2.
I
= 2
π
+
4
3
3.
I
= 2
π

8
3
4.
I
=
π
+
8
3
5.
I
= 4
π

4
3
6.
I
= 4
π
+
4
3
Explanation:
Since
tan
2
x
= sec
2
x

1
,
it follows that
tan
4
x
= tan
2
x
(sec
2
x

1)
= tan
2
x
sec
2

tan
2
x.
Thus, using the same trig identity as before,
we see that
tan
4
x
= (tan
2
x

1) sec
2
x
+ 1
,
in which case
I
= 4
Z
π/
4
0
n
(tan
2
x

1) sec
2
x
+ 1
o
dx.
The whole point of this use of trig identities is
that
d
dx
tan
x
= sec
2
x,
so set
u
= tan
x
. Then
du
= sec
2
xdx,
while
x
= 0
=
⇒
u
= 0
,
x
=
π
4
=
⇒
u
= 1
.
In this case,
I
= 4
Z
1
0
(
u
2

1)
du
+ 4
h
x
i
π/
4
0
= 4
h
1
3
u
3

u
i
1
0
+
π .
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This note was uploaded on 06/17/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Cepparo

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