What is the integral of (6e^{-6x})/(2-e^{-6x)} ?

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

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integral of (6e^{-6x})/(2-e^{-6x)}

\int \frac{6 e ^{- 6 x}}{2 - e ^{- 6 x}} d x

ln 2 - e^{- 6 x} + C

Solution steps

\int \frac{6 e ^{- 6 x}}{2 - e ^{- 6 x}} d x

Take the constant out:

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

= 6 \cdot \int \frac{e ^{- 6 x}}{2 - e ^{- 6 x}} d x

Apply u-substitution

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

= 6 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 2 - e^{- 6 x}

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

Simplify

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

= ln 2 - e^{- 6 x}

Add a constant to the solution

= ln 2 - e^{- 6 x} + C

\int e^{\frac{1}{y}} d y

4 x - 2 x

lo g_{2} (e^{- x} cos (π x))

x \to \infty lim (r^{x})

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