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

\frac{x d y}{d x} = 5 x e^{x} - y + 9 x^{2}

\int 5 x^{2} e^{6 x} d x

\int sec (t) d t

\frac{d y}{d x}, 4 x y = 3

\frac{d}{d t} (- 2 a sin (2 t) + b cos (2 t) \cdot 2)

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