What is the integral of (2e^{3x})/(5-4e^{3x)} ?

The integral of (2e^{3x})/(5-4e^{3x)} is -1/6 ln|5-4e^{3x}|+C

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integral of (2e^{3x})/(5-4e^{3x)}

\int \frac{2 e ^{3 x}}{5 - 4 e ^{3 x}} d x

- \frac{1}{6} ln 5 - 4 e^{3 x} + C

Solution steps

\int \frac{2 e ^{3 x}}{5 - 4 e ^{3 x}} d x

Take the constant out:

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

= 2 \cdot \int \frac{e ^{3 x}}{5 - 4 e ^{3 x}} d x

Apply u-substitution

: \int - \frac{1}{12 u} d u

= 2 \cdot \int - \frac{1}{12 u} d u

Take the constant out:

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

= 2 (- \frac{1}{12} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{12} ln ∣ u ∣)

Substitute back

u = 5 - 4 e^{3 x}

= 2 (- \frac{1}{12} ln 5 - 4 e^{3 x})

Simplify

2 (- \frac{1}{12} ln 5 - 4 e^{3 x}) : - \frac{1}{6} ln 5 - 4 e^{3 x}

= - \frac{1}{6} ln 5 - 4 e^{3 x}

Add a constant to the solution

= - \frac{1}{6} ln 5 - 4 e^{3 x} + C

y^{^{'''}} - 12 y^{^{''}} + 47 y^{^{'}} - 66 y = 0

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What is the integral of (2e^{3x})/(5-4e^{3x)} ?
The integral of (2e^{3x})/(5-4e^{3x)} is -1/6 ln|5-4e^{3x}|+C