What is the integral of 1/(e^{-5x)+2} ?

The integral of 1/(e^{-5x)+2} is x/2+1/10 ln|e^{-5x}+2|+C

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integral of 1/(e^{-5x)+2}

\int \frac{1}{e ^{- 5 x} + 2} d x

\frac{x}{2} + \frac{1}{10} ln e^{- 5 x} + 2 + C

Solution steps

\int \frac{1}{e ^{- 5 x} + 2} d x

Apply u-substitution

: \int - \frac{1}{5 ( e ^{u} + 2 )} d u

= \int - \frac{1}{5 ( e ^{u} + 2 )} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int \frac{1}{e ^{u} + 2} d u

Apply u-substitution

: \int \frac{1}{v ( v + 2 )} d v

= - \frac{1}{5} \cdot \int \frac{1}{v ( v + 2 )} d v

Take the partial fraction of

\frac{1}{v ( v + 2 )} : \frac{1}{2 v} - \frac{1}{2 ( v + 2 )}

= - \frac{1}{5} \cdot \int \frac{1}{2 v} - \frac{1}{2 ( v + 2 )} 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}{5} (\int \frac{1}{2 v} d v - \int \frac{1}{2 ( v + 2 )} d v)

\int \frac{1}{2 v} d v = \frac{1}{2} ln ∣ v ∣

\int \frac{1}{2 ( v + 2 )} d v = \frac{1}{2} ln ∣ v + 2 ∣

= - \frac{1}{5} (\frac{1}{2} ln ∣ v ∣ - \frac{1}{2} ln ∣ v + 2 ∣)

Substitute back

= - \frac{1}{5} (\frac{1}{2} ln e^{- 5 x} - \frac{1}{2} ln e^{- 5 x} + 2)

Simplify

- \frac{1}{5} (\frac{1}{2} ln e^{- 5 x} - \frac{1}{2} ln e^{- 5 x} + 2) : \frac{x}{2} + \frac{1}{10} ln e^{- 5 x} + 2

= \frac{x}{2} + \frac{1}{10} ln e^{- 5 x} + 2

Add a constant to the solution

= \frac{x}{2} + \frac{1}{10} ln e^{- 5 x} + 2 + C

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What is the integral of 1/(e^{-5x)+2} ?
The integral of 1/(e^{-5x)+2} is x/2+1/10 ln|e^{-5x}+2|+C