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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 e^x=t, e^xdx=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

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Sayak

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