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COMEDK Exam 2015 Solved Paper
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© examsnet.com
Question : 162 of 180
Marks:
+1
,
-0
e
∫
1
log
x
dx
=
1
e
−
1
e
+
1
0
Validate
Solution:
We have,
e
∫
1
log
x
dx
=
[
log
x
∫
1
dx
]
1
e
−
e
∫
1
(
d
d
x
log
x
∫
1
dx
)
dx
=
[
log
x
⋅
x
]
1
e
−
e
∫
1
1
x
⋅
x
dx
=
[
e
log
e
−
1
log
(
1
)
]
−
e
∫
1
1
dx
=
e
−
[
x
]
1
e
=
e
−
[
e
−
1
]
=
1
© examsnet.com
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