$\displaystyle\int \frac{dx}{e^{x}+e^{-x}}$ is equal to

Question

Accepted Answer

A. $	an^{-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 $=\int \dfrac{dt}{t^2+1}=	an^{-1}t$.
4. Back-substitute: $	an^{-1}(e^{x})+C$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.350_

Answer

$	an^{-1}(e^{-x})+C$

Answer

$\log(e^{x}-e^{-x})+C$

Answer

$\log(e^{x}+e^{-x})+C$

$\displaystyle\int \frac{dx}{e^{x}+e^{-x}}$ is equal to

Keep practising Straight Lines and Conic Sections