(B) Let $I = \int \frac{dx}{e^x + e^{-x}}$.Multiply the numerator and denominator by $e^x$:$I = \int \frac{e^x}{e^{2x} + 1} dx$.Let $e^x = t$. Then $e

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(B) Let $I = \int \frac{dx}{e^x + e^{-x}}$.Multiply the numerator and denominator by $e^x$:$I = \int \frac{e^x}{e^{2x} + 1} dx$.Let $e^x = t$. Then $e^x dx = dt$.Substituting these into the integral:$I = \int \frac{dt}{t^2 + 1}$.The integral of $\frac{1}{t^2 + 1}$ is $\tan^{-1}(t) + C$.Thus,$I = \tan^{-1}(e^x) + C$.