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

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

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

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

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

Solution steps

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

Simplify

\frac{e ^{3 x}}{e ^{3 x} - e ^{- 3 x}} : \frac{e ^{6 x}}{e ^{6 x} - 1}

= \int \frac{e ^{6 x}}{e ^{6 x} - 1} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = e^{6 x} - 1

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

Add a constant to the solution

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

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\int 9 sec^{3} (x) d x

\frac{d ^{2}}{d x ^{2}} (1 - x^{2})

\int 23 e^{x} d x

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