IIT JEE Advanced 2008 Paper 2

Section: Mathematics

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Question : 48 of 66

Marks: +1, -0

Let I = ∫

ex

e4x+e2x+1

dx, J = ∫

e−x

e−4x+e−2x+1

dx. Then, for an arbitrary constant C, the value of J - 1 equals

[JEE Adv 2008 P2]

1
2
‌log(
e4x−e2x+1
e4x+e2x+1
) + C
1
2
‌log(
e2x+ex+1
e2x−ex+1
) + C
1
2
‌log(
e2x−ex+1
e2x+ex+1
) + C
1
2
‌log(
e4x+e2x+1
e4x−e2x+1
) + C

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