What is the integral of 2/(e^{-x)+1} ?

The integral of 2/(e^{-x)+1} is 2(ln|e^{-x}+1|+x)+C

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integral of 2/(e^{-x)+1}

\int \frac{2}{e ^{- x} + 1} d x

2 (ln e^{- x} + 1 + x) + C

Solution steps

\int \frac{2}{e ^{- x} + 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{e ^{- x} + 1} d x

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

= 2 \cdot \int - \frac{e ^{- x}}{e ^{- x} + 1} + 1 d x

Apply the Sum Rule:

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

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

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

\int 1 d x = x

= 2 (- (- ln e^{- x} + 1) + x)

Simplify

= 2 (ln e^{- x} + 1 + x)

Add a constant to the solution

= 2 (ln e^{- x} + 1 + x) + C

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What is the integral of 2/(e^{-x)+1} ?
The integral of 2/(e^{-x)+1} is 2(ln|e^{-x}+1|+x)+C