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How do you integrate the following?
1/(e^2x + e^-2x)
Thanks!
2 Answers
- ?Lv 77 years ago
integral [1/(e^2x + e^-2x)] dx
= integral [e^2x/((e^2x)^2 + 1)] dx
= (1/2) integral [2e^2x/((e^2x)^2 + 1)] dx
= (1/2) integral [1/(u^2 + 1)] du , using u = e^2x, du = 2e^2x dx
= (1/2) arctan u + C
= (1/2) arctan(e^2x) + C.
May Jesus richly bless you today!
- cidyahLv 77 years ago
∫ 1/( e^2x + e^-2x) dx
Let u = e^2x
du = 2 e^2x dx
du = 2 u dx
dx = 1 / 2u du
∫ 1/( e^2x + e^-2x) dx = (1/2) ∫ 1 /u (u+ 1/u) du
= (1/2) ∫ 1 / (u^2+1) du
=(1/2) tan^-1(u)
= (1/2) tan^-1( e^2x) + C
