What is the integral of 1/(e^{-v)+1} ?

The integral of 1/(e^{-v)+1} is ln|e^{-v}+1|+v+C

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integral of 1/(e^{-v)+1}

\int \frac{1}{e ^{- v} + 1} d v

ln e^{- v} + 1 + v + C

Solution steps

\int \frac{1}{e ^{- v} + 1} d v

\frac{1}{e ^{- v} + 1} = - \frac{e ^{- v}}{e ^{- v} + 1} + 1

= \int - \frac{e ^{- v}}{e ^{- v} + 1} + 1 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{e ^{- v}}{e ^{- v} + 1} d v + \int 1 d v

\int \frac{e ^{- v}}{e ^{- v} + 1} d v = - ln e^{- v} + 1

\int 1 d v = v

= - (- ln e^{- v} + 1) + v

Simplify

= ln e^{- v} + 1 + v

Add a constant to the solution

= ln e^{- v} + 1 + v + C

\int x^{3} \cdot sin (2 x) d x

y = \frac{5}{x ^{4}} - \frac{7}{x}

\frac{3 s}{( s - 1 ) ( s ^{2} + 1 )}

x \to \infty lim (2 x - 2)

\frac{1}{4 s + 1}

What is the integral of 1/(e^{-v)+1} ?
The integral of 1/(e^{-v)+1} is ln|e^{-v}+1|+v+C