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Choose the correct answers in Exercises 41 to 44.

    Q41.    \int\frac{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

Answers (1)

best_answer

\int\frac{dx}{e^x + e^{-x}} 

the above integral can be re arranged as

\\=\int \frac{e^{x}}{e^{2x}+1}dx\\

let ex=t, exdx=dt

\int\frac{dx}{e^x + e^{-x}} 

\\=\int \frac{1}{t^{2}+1}dt\\ =tan^{-1}t+c\\ =tan^{-1}(e^{x})+c

(A) is correct

Posted by

Sayak

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