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

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

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

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

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

Solution steps

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

Simplify

\frac{e ^{x}}{e ^{x} + e ^{- x}} : \frac{e ^{2 x}}{e ^{2 x} + 1}

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = e^{2 x} + 1

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

Add a constant to the solution

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

n = 0 \sum \infty \frac{1}{2 ^{2^{n}}}

y = π^{3 x + 1}

\frac{d z}{d z}

\frac{d}{d x} ((x + y + 1)^{2})

f (x) = cos (a^{9} + x^{9})

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