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

The integral of (e^{6x})/(e^{6x)-3e^{-6x}} is 1/12 ln|e^{12x}-3|+C

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

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

\frac{1}{12} ln e^{12 x} - 3 + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{2 u}}{6 ( e ^{2 u} - 3 )} d u

= \int \frac{e ^{2 u}}{6 ( e ^{2 u} - 3 )} 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{e ^{2 u}}{e ^{2 u} - 3} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

= \frac{1}{6} \cdot \int \frac{1}{2 v} d v

Take the constant out:

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

= \frac{1}{6} \cdot \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{6} \cdot \frac{1}{2} ln ∣ v ∣

Substitute back

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

Simplify

\frac{1}{6} \cdot \frac{1}{2} ln e^{2 \cdot 6 x} - 3 : \frac{1}{12} ln e^{12 x} - 3

= \frac{1}{12} ln e^{12 x} - 3

Add a constant to the solution

= \frac{1}{12} ln e^{12 x} - 3 + C

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What is the integral of (e^{6x})/(e^{6x)-3e^{-6x}} ?
The integral of (e^{6x})/(e^{6x)-3e^{-6x}} is 1/12 ln|e^{12x}-3|+C