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

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

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

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

arctan (e^{x}) + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= arctan (u)

Substitute back

u = e^{x}

= arctan (e^{x})

Add a constant to the solution

= arctan (e^{x}) + C

t \to 0 + lim (\frac{4}{t} - \frac{1}{e ^{t} - 1})

\frac{d}{d x} (2 csc (x) cot (x))

y = 2 x + 1, at x = \frac{1}{4}

\frac{s ^{2}}{( s - 3 ) ( s ^{2} + 5 )}

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

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