What is the integral of 1/(sqrt(e^{2x)-1)} ?

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

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integral of 1/(sqrt(e^{2x)-1)}

\int \frac{1}{e ^{2 x} - 1} d x

arctan (e^{2 x} - 1) + C

Solution steps

\int \frac{1}{e ^{2 x} - 1} 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^{2 x} - 1

= arctan (e^{2 x} - 1)

Add a constant to the solution

= arctan (e^{2 x} - 1) + C

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

\frac{d}{d y} (ln (\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1}))

ln (1 + 3 x)

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{4 x})

y^{^{'}} = 2 + 3 x + 8 y + 12 x y

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