# calc

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integrate 1/(1+(e^-x))

• calc -

1/(1+(e^-x))
= 1/(1 + 1/e^x)
= 1/( (e^x + 1)/e^x )
= e^x/(e^x + 1)

now by "recognition" of how to differentiate Ln(anything) = anything'/anything
I get
∫1/(1+(e^-x)) dx
= ln(e^x + 1) + c

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