What is the integral of (e^{2x})/(sqrt(e^{4x)-1)} ?

The integral of (e^{2x})/(sqrt(e^{4x)-1)} is 1/2 ln|sqrt(e^{4x)-1}+e^{2x}|+C

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integral of (e^{2x})/(sqrt(e^{4x)-1)}

\int \frac{e ^{2 x}}{e ^{4 x} - 1} d x

\frac{1}{2} ln e^{4 x} - 1 + e^{2 x} + C

Solution steps

\int \frac{e ^{2 x}}{e ^{4 x} - 1} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} - 1} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} ln tan (arcsec (e^{2 x})) + sec (arcsec (e^{2 x}))

Simplify

\frac{1}{2} ln tan (arcsec (e^{2 x})) + sec (arcsec (e^{2 x})) : \frac{1}{2} ln e^{4 x} - 1 + e^{2 x}

= \frac{1}{2} ln e^{4 x} - 1 + e^{2 x}

Add a constant to the solution

= \frac{1}{2} ln e^{4 x} - 1 + e^{2 x} + C

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What is the integral of (e^{2x})/(sqrt(e^{4x)-1)} ?
The integral of (e^{2x})/(sqrt(e^{4x)-1)} is 1/2 ln|sqrt(e^{4x)-1}+e^{2x}|+C