KEAM 2011 Math Paper

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Question : 61 of 120

Marks: +1, -0

∫

4ex+6e−x

9ex−4e−x

dx is equal to

3
2
x+
35
36
‌log|9ex−4|+C
3
2
x−
35
36
‌log|9e2x−4|+C
−
3
2
x+
35
36
‌log|9e2x−4|+C
−
5
2
x+
35
36
‌log|9e2x−4|+C
5
2
x+
35
36
‌log|9e2x−4|+C

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