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MATH 112SPRING 2008SECTION 5
Achilleas Sinefakopoulos
asin@math.cornell.edu
1. Find the error in the proof of the following claim.
Claim
: 0 = 1.
Proof
. By integration by parts (with
dv
=
dx
and
u
=
1
x
), we get
Z
1
x
dx
=
x
·
1
x

Z
x
±

1
x
2
¶
dx
= 1 +
Z
1
x
dx.
So by cancelling
Z
1
x
dx
from both sides, we get 0 = 1.
Solution
. Cancelling
Z
1
x
dx
is not allowed because it denotes a“family” of functions
whose derivatives is
1
x
and not just one function. That’s why we include “ +
C
” in an
indeﬁnite integral.
2. Find the following
indeﬁnite
integrals.
(i)
Z
x
ln
xdx
=
x
2
2
ln
x

x
2
4
+
C
Hint
: Apply integration by parts with
u
= ln
x
and
dv
=
xdx
.
(ii)
Z
x
sin

1
xdx
=
x
2
2
sin

1
x

1
4
sin

1
x
+
x
4
√
1

x
2
+
C
Hint
: Apply integration by parts with
u
= sin

1
x
and
dv
=
xdx
.
(iii)
Z
(ln
x
)
2
dx
=
x
(ln
x
)
2

2
x
ln
x
+ 2
x
+
C
Hint
: Apply integration by parts with
u
= (ln
x
)
2
and
dv
=
dx
. Then apply
integration by parts again to ﬁnd
Z
ln
xdx
.
(iv)
Z
sin
3
xdx
=

1
3
sin
2
x
cos
x

2
3
cos
x
+
C
(v)
Z
ln(
x
+
√
1 +
x
2
)
dx
=
x
ln(
x
+
√
1 +
x
2
)

√
1 +
x
2
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This note was uploaded on 02/28/2010 for the course MATH 1120 taught by Professor Gross during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 GROSS

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