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))}

integral

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

arctan (e^{x}) + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (u)

Substitute back

u = e^{x}

= arctan (e^{x})

Add a constant to the solution

= arctan (e^{x}) + C

\frac{sin ^{2} ( x )}{( 1 + cos ( x ) ) ^{2}}

cos^{6} (x)

\frac{1}{a ^{x}}

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

\frac{x ^{2}}{( x ^{3} + 1 )}

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