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

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

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

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

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

Solution steps

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

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 y} - 1

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

Add a constant to the solution

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

\int_{0}^{\frac{π}{4}} x^{2} sin (2 x) d x

\frac{d}{d x} (cos (x) + ln (cos (x)) + x sin (x))

\int \frac{x ^{3}}{1 + x ^{3}} d x

x \to 1 lim (x^{2} + 2 x - 2)

f (x) = 4 e^{x} - x^{16} x, at x = 0

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