KEAM 2015 Math Paper

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Question : 8 of 120

Marks: +1, -0

∫

4ex

2ex−5e−x

dx=

4‌log|ex−5|+C
1
4
‌log|e2x−5|+C
log|ex−5e−x|+C
4‌log|e2x−5|+C
log|2e2x−5|+C

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