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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Use the common integral:

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

= \frac{1}{2} arcsin (u)

Substitute back

u = e^{2 x}

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

Add a constant to the solution

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

x \to 2 lim (\frac{x + 1}{x ^{2} - 4})

\frac{d}{d x} ((x + x)^{2})

a re a y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{6}

\int \frac{1}{- x ^{2} - a + b} d x

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

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