Maybe Simplify First?

Calculus Level 1

Evaluate the indefinite integral below.

∫ d x 1 + e − x \large \int \frac{dx}{1 + e^{-x}}

ln ⁡ ( e x + 1 ) \ln(e^x + 1) + C ln ⁡ ( e x − 1 ) \ln(e^x - 1) + C ln ⁡ ( e − x + 1 ) \ln(e^{-x} + 1) + C ln ⁡ ( e − x − 1 ) \ln(e^{-x} - 1) + C

Andrew Ellinor by Andrew Ellinor , 33, USA

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1 solution

Atul Shivam
Oct 6, 2015

The above integrand can also be mentioned as e x 1 + e x \frac {e^x}{1 + e^x } and hence substitute 1 + e x 1+e^x equal to ”t" then the whole integrand will convert into 1 t d t \frac {1}{ t } dt and it has simple integration of ln(t) +c hence substitute t= 1 + e x 1+ e^x to get the result

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