JEE Main 12 Jan 2019 Shift 2 Solved Paper

Section: Mathematics

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Question : 44 of 90

Marks: +1, -0

The integral

e

∫

1

{(

x

e

)2x−(

e

x

)x}loge xdx is equal to :

[12 Jan 2019 Shift 2]

1
2
−e−
1
22
3
2
−
1
e
−
1
2e2
−
1
2
+
1
e
−
1
2e2
3
2
−e−
1
2e2

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