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Question
Mathematics
The value of the integral displaystyle ∫1e (1+ log x/3x) dx is equal to
Q. The value of the integral
∫
1
e
3
x
1
+
l
o
g
x
d
x
is equal to
2843
231
KEAM
KEAM 2014
Integrals
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A
4
1
B
2
1
C
4
3
D
e
E
e
1
Solution:
Let
I
=
1
∫
e
3
x
1
+
l
o
g
x
d
x
=
1
∫
e
3
x
1
d
x
+
3
1
1
∫
e
x
l
o
g
x
d
x
Put
lo
g
x
=
t
⇒
x
1
d
x
=
d
t
∴
I
=
1
∫
e
3
x
1
d
x
+
3
1
∫
0
1
t
d
t
=
[
3
1
lo
g
x
]
e
+
3
1
[
2
t
2
]
0
1
=
3
1
[
lo
g
e
−
lo
g
1
]
+
6
1
[
1
2
−
0
]
=
3
1
[
1
−
0
]
+
6
1
=
6
3
=
2
1