What is the integral of (e^{3x})/(4-9e^{6x)} ?

The integral of (e^{3x})/(4-9e^{6x)} is 1/36 (ln| 3/2 e^{3x}+1|-ln| 3/2 e^{3x}-1|)+C

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integral of (e^{3x})/(4-9e^{6x)}

\int \frac{e ^{3 x}}{4 - 9 e ^{6 x}} d x

\frac{1}{36} (ln \frac{3}{2} e^{3 x} + 1 - ln \frac{3}{2} e^{3 x} - 1) + C

Solution steps

\int \frac{e ^{3 x}}{4 - 9 e ^{6 x}} d x

Apply u-substitution

: \int \frac{1}{3 ( 4 - 9 u ^{2} )} d u

= \int \frac{1}{3 ( 4 - 9 u ^{2} )} 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{1}{4 - 9 u ^{2}} d u

Apply Integral Substitution

= \frac{1}{3} \cdot \int \frac{1}{6 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{6} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{3} \cdot \frac{1}{6} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{3} \cdot \frac{1}{6} (\frac{ln \frac{3}{2} e ^{3 x} + 1}{2} - \frac{ln \frac{3}{2} e ^{3 x} - 1}{2})

Simplify

\frac{1}{3} \cdot \frac{1}{6} (\frac{ln \frac{3}{2} e ^{3 x} + 1}{2} - \frac{ln \frac{3}{2} e ^{3 x} - 1}{2}) : \frac{1}{36} (ln \frac{3}{2} e^{3 x} + 1 - ln \frac{3}{2} e^{3 x} - 1)

= \frac{1}{36} (ln \frac{3}{2} e^{3 x} + 1 - ln \frac{3}{2} e^{3 x} - 1)

Add a constant to the solution

= \frac{1}{36} (ln \frac{3}{2} e^{3 x} + 1 - ln \frac{3}{2} e^{3 x} - 1) + C

y^{^{'}} - \frac{y}{x} = y^{2}

x y (\frac{d y}{d x}) = 3

\int_{1}^{4} \frac{5 ln ( x )}{x ^{2}} d x

\frac{d}{d x} (1 - e^{- x} - x e^{- x})

t y^{^{'}} = y - 5 t^{2} sin (3 t), y (\frac{π}{3}) = 0

What is the integral of (e^{3x})/(4-9e^{6x)} ?
The integral of (e^{3x})/(4-9e^{6x)} is 1/36 (ln| 3/2 e^{3x}+1|-ln| 3/2 e^{3x}-1|)+C