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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

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