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

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

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

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

ln e^{x} + e^{- x} + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = e^{x} + e^{- x}

= ln e^{x} + e^{- x}

Add a constant to the solution

= ln e^{x} + e^{- x} + C

\int \frac{6 x}{( 4 - 3 x ^{2} ) ^{4}} d x

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

x^{3} - 6 x^{2} - 15 x + 4

x \to 5 lim (x^{- 2})

y^{^{'}} = \frac{9 x}{y + x ^{2} y}

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