# Indefinite integral

How do I find the indefinite integral of 1/[(e^x)+(e^-x)] ???

Do I have to multiply by [(e^x)-(e^-x)]/[(e^x)-(e^-x)] to eliminate the denominator???? !

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are you trying to do:
$$\int\frac{dx}{e^x+e^{-x}}$$

if so, for a=b=c:
$$\int\frac{dx}{e^{ax}+e^{-bx}}=\int\frac{dx}{e^{cx}+e^{-cx}}=\frac{tan^{-1}(e^{cx})}{c}+C$$
i hope that helps

p53ud0 dr34m5 said:
are you trying to do:
$$\int\frac{dx}{e^x+e^{-x}}$$

if so, for a=b=c:
$$\int\frac{dx}{e^{ax}+e^{-bx}}=\int\frac{dx}{e^{cx}+e^{-cx}}=\frac{tan^{-1}(e^{cx})}{c}+C$$
i hope that helps

I dont get it

$$\frac{1}{e^x + e^{-x}} = \frac{1}{\frac{(e^x)^2+1}{e^x}}$$

Can you go from there?

James R
$$\frac{e^x + e^{-x}}{2} = \cosh x$$