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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{3 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{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = e^{3 x} - 4

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

Add a constant to the solution

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

\int \frac{x ^{4} + x + 4}{x ^{2} + 2} d x

4^{x}

\int \frac{11 x ^{2} + 5 x - 4}{x ^{3} + x ^{2} - 2 x} d x

\int \frac{ln ( x ) ln ( x )}{x} d x

\int \frac{x ^{3}}{x ^{8} - 4} d x

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