1/(e^x+4e^-x) using the substitution u=e^x. I got 1/4e^x - e^-x + c ,
however the answers in my textbook say 1/2tan^-1(1/2e^x + c).
Thanks in advance for your time and help
however the answers in my textbook say 1/2tan^-1(1/2e^x + c).
Thanks in advance for your time and help
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Let u = e^x, du = e^x dx = u dx.
So, ∫ dx/(e^x + 4e^(-x))
= ∫ (du/u) / (u + 4/u)
= ∫ du / (u^2 + 4)
= (1/2) arctan(u/2) + C
= (1/2) arctan((1/2)e^x) + C.
I hope this helps!
So, ∫ dx/(e^x + 4e^(-x))
= ∫ (du/u) / (u + 4/u)
= ∫ du / (u^2 + 4)
= (1/2) arctan(u/2) + C
= (1/2) arctan((1/2)e^x) + C.
I hope this helps!
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Just confused about the transition between steps 2 (∫ du / (u^2 + 4)) and 3 (= (1/2) arctan(u/2) + C) - when the integration is being performed. Could someone please explain how you get the result in step 3? Thanks
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