$\displaystyle\int \frac{dx}{e^{x}+e^{-x}}$ is equal to
A$\tan^{-1}(e^{x})+C$
B$\tan^{-1}(e^{-x})+C$
C$\log(e^{x}-e^{-x})+C$
D$\log(e^{x}+e^{-x})+C$
Answer & Solution
Correct answer: A. $\tan^{-1}(e^{x})+C$
1. Multiply numerator and denominator by $e^{x}$: $\dfrac{e^{x}}{e^{2x}+1}$.
2. Put $t=e^{x}$, $dt=e^{x}\,dx$.
3. Integral $=\int \dfrac{dt}{t^2+1}=\tan^{-1}t$.
4. Back-substitute: $\tan^{-1}(e^{x})+C$.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.350_
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