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

The integral of (e^{3x})/(sqrt(4-e^{6x))} is 1/3 arcsin(1/2 e^{3x})+C

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

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

\frac{1}{3} arcsin (\frac{1}{2} e^{3 x}) + C

Solution steps

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

Apply u-substitution

= \int \frac{e ^{u}}{3 4 - e ^{2 u}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{e ^{u}}{4 - e ^{2 u}} d u

Apply u-substitution

= \frac{1}{3} \cdot \int \frac{1}{4 - v ^{2}} d v

Apply Trigonometric Substitution

= \frac{1}{3} \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot w

Substitute back

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

Simplify

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

Add a constant to the solution

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

What is the integral of (e^{3x})/(sqrt(4-e^{6x))} ?
The integral of (e^{3x})/(sqrt(4-e^{6x))} is 1/3 arcsin(1/2 e^{3x})+C