What is the integral of (e^{2x})/(sqrt(16-9e^{4x))} ?

The integral of (e^{2x})/(sqrt(16-9e^{4x))} is 1/6 arcsin(3/4 e^{2x})+C

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integral of (e^{2x})/(sqrt(16-9e^{4x))}

\int \frac{e ^{2 x}}{16 - 9 e ^{4 x}} d x

\frac{1}{6} arcsin (\frac{3}{4} e^{2 x}) + C

Solution steps

\int \frac{e ^{2 x}}{16 - 9 e ^{4 x}} d x

Apply u-substitution

: \int \frac{e ^{u}}{2 16 - 9 e ^{2 u}} d u

= \int \frac{e ^{u}}{2 16 - 9 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}{2} \cdot \int \frac{e ^{u}}{16 - 9 e ^{2 u}} d u

Apply u-substitution

: \int \frac{1}{16 - 9 v ^{2}} d v

= \frac{1}{2} \cdot \int \frac{1}{16 - 9 v ^{2}} d v

Apply Trigonometric Substitution

: \int \frac{1}{3} d w

= \frac{1}{2} \cdot \int \frac{1}{3} d w

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 1 d w

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot w

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{4} e^{2 x}) : \frac{1}{6} arcsin (\frac{3}{4} e^{2 x})

= \frac{1}{6} arcsin (\frac{3}{4} e^{2 x})

Add a constant to the solution

= \frac{1}{6} arcsin (\frac{3}{4} e^{2 x}) + C

\frac{d}{d x} (2 cos (x) - 4 x sin (x) - x^{2} cos (x))

\int \frac{1}{3 s h x + 5 c h x} d x

\frac{5 s}{( s ^{2} + 4 ) ( s + 1 )}

\frac{d y}{d x} = 0.002 y (500 - y) - 80

\frac{d}{d x} (x - 16)

What is the integral of (e^{2x})/(sqrt(16-9e^{4x))} ?
The integral of (e^{2x})/(sqrt(16-9e^{4x))} is 1/6 arcsin(3/4 e^{2x})+C