What is the integral of 1/(3+e^{-2x)} ?

The integral of 1/(3+e^{-2x)} is x/3+1/6 ln|e^{-2x}+3|+C

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integral of 1/(3+e^{-2x)}

\int \frac{1}{3 + e ^{- 2 x}} d x

\frac{x}{3} + \frac{1}{6} ln e^{- 2 x} + 3 + C

Solution steps

\int \frac{1}{3 + e ^{- 2 x}} d x

Apply u-substitution

: \int - \frac{1}{2 ( 3 + e ^{u} )} d u

= \int - \frac{1}{2 ( 3 + e ^{u} )} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{3 + e ^{u}} d u

Apply u-substitution

: \int \frac{1}{v ( 3 + v )} d v

= - \frac{1}{2} \cdot \int \frac{1}{v ( 3 + v )} d v

Take the partial fraction of

\frac{1}{v ( 3 + v )} : \frac{1}{3 v} - \frac{1}{3 ( v + 3 )}

= - \frac{1}{2} \cdot \int \frac{1}{3 v} - \frac{1}{3 ( v + 3 )} 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}{2} (\int \frac{1}{3 v} d v - \int \frac{1}{3 ( v + 3 )} d v)

\int \frac{1}{3 v} d v = \frac{1}{3} ln ∣ v ∣

\int \frac{1}{3 ( v + 3 )} d v = \frac{1}{3} ln ∣ v + 3 ∣

= - \frac{1}{2} (\frac{1}{3} ln ∣ v ∣ - \frac{1}{3} ln ∣ v + 3 ∣)

Substitute back

= - \frac{1}{2} (\frac{1}{3} ln e^{- 2 x} - \frac{1}{3} ln e^{- 2 x} + 3)

Simplify

- \frac{1}{2} (\frac{1}{3} ln e^{- 2 x} - \frac{1}{3} ln e^{- 2 x} + 3) : \frac{x}{3} + \frac{1}{6} ln e^{- 2 x} + 3

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

Add a constant to the solution

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

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What is the integral of 1/(3+e^{-2x)} ?
The integral of 1/(3+e^{-2x)} is x/3+1/6 ln|e^{-2x}+3|+C