What is the integral of (e^{2x})/(sqrt(9-4e^{4x))} ?

The integral of (e^{2x})/(sqrt(9-4e^{4x))} is 1/4 arcsin(2/3 e^{2x})+C

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integral of (e^{2x})/(sqrt(9-4e^{4x))}

\int \frac{e ^{2 x}}{9 - 4 e ^{4 x}} d x

\frac{1}{4} arcsin (\frac{2}{3} e^{2 x}) + C

Solution steps

\int \frac{e ^{2 x}}{9 - 4 e ^{4 x}} d x

Apply u-substitution

: \int \frac{e ^{\frac{u}{2}}}{4 9 - 4 e ^{u}} d u

= \int \frac{e ^{\frac{u}{2}}}{4 9 - 4 e ^{u}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{e ^{\frac{u}{2}}}{9 - 4 e ^{u}} d u

Apply u-substitution

: \int \frac{2 e ^{v}}{9 - 4 e ^{2 v}} d v

= \frac{1}{4} \cdot \int \frac{2 e ^{v}}{9 - 4 e ^{2 v}} d v

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{e ^{v}}{9 - 4 e ^{2 v}} d v

Apply u-substitution

: \int \frac{1}{9 - 4 w ^{2}} d w

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{9 - 4 w ^{2}} d w

Apply Trigonometric Substitution

: \int \frac{1}{2} d t

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{2} d t

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \frac{1}{2} \cdot \int 1 d t

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot t

Substitute back

= \frac{1}{4} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{3} e^{\frac{4 x}{2}})

Simplify

\frac{1}{4} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{3} e^{\frac{4 x}{2}}) : \frac{1}{4} arcsin (\frac{2}{3} e^{2 x})

= \frac{1}{4} arcsin (\frac{2}{3} e^{2 x})

Add a constant to the solution

= \frac{1}{4} arcsin (\frac{2}{3} e^{2 x}) + C

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What is the integral of (e^{2x})/(sqrt(9-4e^{4x))} ?
The integral of (e^{2x})/(sqrt(9-4e^{4x))} is 1/4 arcsin(2/3 e^{2x})+C