What is the integral of (1-e^{-4x})/(1+e^{-4x)} ?

The integral of (1-e^{-4x})/(1+e^{-4x)} is x+1/2 ln|e^{-4x}+1|+C

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integral of (1-e^{-4x})/(1+e^{-4x)}

\int \frac{1 - e ^{- 4 x}}{1 + e ^{- 4 x}} d x

x + \frac{1}{2} ln e^{- 4 x} + 1 + C

Solution steps

\int \frac{1 - e ^{- 4 x}}{1 + e ^{- 4 x}} d x

Expand

\frac{1 - e ^{- 4 x}}{1 + e ^{- 4 x}} : \frac{1}{1 + e ^{- 4 x}} - \frac{e ^{- 4 x}}{1 + e ^{- 4 x}}

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{1}{1 + e ^{- 4 x}} d x - \int \frac{e ^{- 4 x}}{1 + e ^{- 4 x}} d x

\int \frac{1}{1 + e ^{- 4 x}} d x = \frac{1}{4} ln 1 + e^{- 4 x} + x

\int \frac{e ^{- 4 x}}{1 + e ^{- 4 x}} d x = - \frac{1}{4} ln 1 + e^{- 4 x}

= \frac{1}{4} ln 1 + e^{- 4 x} + x - (- \frac{1}{4} ln 1 + e^{- 4 x})

Simplify

\frac{1}{4} ln 1 + e^{- 4 x} + x - (- \frac{1}{4} ln 1 + e^{- 4 x}) : x + \frac{1}{2} ln e^{- 4 x} + 1

= x + \frac{1}{2} ln e^{- 4 x} + 1

Add a constant to the solution

= x + \frac{1}{2} ln e^{- 4 x} + 1 + C

d er i v a t i v e (1 + 4 x)^{\frac{3}{2}}

\frac{d}{d x} (\frac{1}{8 x ^{- 4}} + \frac{1}{4 x ^{2}})

\frac{d}{d x} (2 x - cos (x))

2 y^{^{'}} = 3 t + 2

\frac{d}{d x} (e^{- \frac{π}{2}})

What is the integral of (1-e^{-4x})/(1+e^{-4x)} ?
The integral of (1-e^{-4x})/(1+e^{-4x)} is x+1/2 ln|e^{-4x}+1|+C