What is the integral of 1/(6+e^{-8x)} ?

The integral of 1/(6+e^{-8x)} is -1/8 (-4/3 x-1/6 ln|e^{-8x}+6|)+C

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integral of 1/(6+e^{-8x)}

\int \frac{1}{6 + e ^{- 8 x}} d x

- \frac{1}{8} (- \frac{4}{3} x - \frac{1}{6} ln e^{- 8 x} + 6) + C

Solution steps

\int \frac{1}{6 + e ^{- 8 x}} d x

Apply u-substitution

: \int - \frac{1}{8 ( 6 + e ^{u} )} d u

= \int - \frac{1}{8 ( 6 + e ^{u} )} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int \frac{1}{6 + e ^{u}} d u

Apply u-substitution

: \int \frac{1}{v ( 6 + v )} d v

= - \frac{1}{8} \cdot \int \frac{1}{v ( 6 + v )} d v

Take the partial fraction of

\frac{1}{v ( 6 + v )} : \frac{1}{6 v} - \frac{1}{6 ( v + 6 )}

= - \frac{1}{8} \cdot \int \frac{1}{6 v} - \frac{1}{6 ( v + 6 )} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{8} (\int \frac{1}{6 v} d v - \int \frac{1}{6 ( v + 6 )} d v)

\int \frac{1}{6 v} d v = \frac{1}{6} ln ∣ v ∣

\int \frac{1}{6 ( v + 6 )} d v = \frac{1}{6} ln ∣ v + 6 ∣

= - \frac{1}{8} (\frac{1}{6} ln ∣ v ∣ - \frac{1}{6} ln ∣ v + 6 ∣)

Substitute back

= - \frac{1}{8} (\frac{1}{6} ln e^{- 8 x} - \frac{1}{6} ln e^{- 8 x} + 6)

\frac{1}{6} ln e^{- 8 x} = - \frac{4}{3} x

= - \frac{1}{8} (- \frac{4}{3} x - \frac{1}{6} ln e^{- 8 x} + 6)

Add a constant to the solution

= - \frac{1}{8} (- \frac{4}{3} x - \frac{1}{6} ln e^{- 8 x} + 6) + C

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What is the integral of 1/(6+e^{-8x)} ?
The integral of 1/(6+e^{-8x)} is -1/8 (-4/3 x-1/6 ln|e^{-8x}+6|)+C