What is the integral of (e^{3x})/(sqrt(4-4e^{6x))} ?

The integral of (e^{3x})/(sqrt(4-4e^{6x))} is 1/6 arcsin(e^{3x})+C

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integral of (e^{3x})/(sqrt(4-4e^{6x))}

\int \frac{e ^{3 x}}{4 - 4 e ^{6 x}} d x

\frac{1}{6} arcsin (e^{3 x}) + C

Solution steps

\int \frac{e ^{3 x}}{4 - 4 e ^{6 x}} d x

4 - 4 e^{6 x} = 2 1 - e^{6 x}

= \int \frac{e ^{3 x}}{2 1 - e ^{6 x}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{e ^{3 x}}{1 - e ^{6 x}} d x

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{1}{3 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 \frac{1}{3} \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} \cdot \frac{1}{3} arcsin (u)

Substitute back

u = e^{3 x}

= \frac{1}{2} \cdot \frac{1}{3} arcsin (e^{3 x})

Simplify

\frac{1}{2} \cdot \frac{1}{3} arcsin (e^{3 x}) : \frac{1}{6} arcsin (e^{3 x})

= \frac{1}{6} arcsin (e^{3 x})

Add a constant to the solution

= \frac{1}{6} arcsin (e^{3 x}) + C

What is the integral of (e^{3x})/(sqrt(4-4e^{6x))} ?
The integral of (e^{3x})/(sqrt(4-4e^{6x))} is 1/6 arcsin(e^{3x})+C