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

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

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

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

- \frac{1}{2} ln 1 - e^{2 y} + y + C

Solution steps

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

\frac{1}{1 - e ^{2 y}} = \frac{e ^{2 y}}{1 - e ^{2 y}} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d y = y

= - \frac{1}{2} ln 1 - e^{2 y} + y

Add a constant to the solution

= - \frac{1}{2} ln 1 - e^{2 y} + y + C

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