Misc 41
Last updated at March 11, 2017 by Teachoo
Transcript
Misc 41 Choose the correct answer ∫▒dx/(e^x + e^(-x) ) is equal to (A) tan^(-1) (e^x )+C (B) tan^(-1)〖(e^(-x) )+C〗 (C) log(e^x-e^(-x) )+C (D) log(e^x+e^(-x) )+C ∫▒dx/(e^x + e^(-x) ) = ∫▒dx/(e^x + 1/e^x ) = ∫1▒(e^x dx)/(e^2x + 1) Let e^x=t dt/dx=e^x dt = e^x dx Substituting, = ∫1▒dt/(t^2 +1) = 〖tan〗^(-1) (t)+ C Putting value of t = 〖tan〗^(-1) (e^x )+ C Hence, answer is (A).
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