What is the integral of 1/((k+e^{-t))} ?

The integral of 1/((k+e^{-t))} is 1/k (ln|e^{-t} 2/k+2|-ln(2/(|k|))+t)+C

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integral of 1/((k+e^{-t))}

integral

\frac{1}{( k + e ^{- t} )}

\frac{1}{k} (ln e^{- t} \frac{2}{k} + 2 - ln (\frac{2}{∣ k ∣}) + t) + C

Solution steps

\int \frac{1}{k + e ^{- t}} d t

Apply u-substitution

= \int - \frac{1}{u ( k + u )} d u

Take the constant out:

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

= - \int \frac{1}{u ( k + u )} d u

Complete the square

u (k + u) : (u + \frac{k}{2})^{2} - \frac{k ^{2}}{4}

= - \int \frac{1}{( u + \frac{k}{2} ) ^{2} - \frac{k ^{2}}{4}} d u

Apply u-substitution

= - \int \frac{4}{4 v ^{2} - k ^{2}} d v

Take the constant out:

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

= - 4 \cdot \int \frac{1}{4 v ^{2} - k ^{2}} d v

Factor

4 v^{2} - k^{2} : - (- 4 v^{2} + k^{2})

= - 4 \cdot \int \frac{1}{- ( - 4 v ^{2} + k ^{2} )} d v

Take the constant out:

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

= - 4 (- \int \frac{1}{- 4 v ^{2} + k ^{2}} d v)

Apply Integral Substitution

= - 4 (- \int \frac{1}{2 k ( - w ^{2} + 1 )} d w)

Take the constant out:

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

= - 4 (- \frac{1}{2 k} \cdot \int \frac{1}{- w ^{2} + 1} d w)

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= - 4 (- \frac{1}{2 k} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

Substitute back

= - 4 (- \frac{1}{2 k} (\frac{ln \frac{2}{k} ( e ^{- t} + \frac{k}{2} ) + 1}{2} - \frac{ln \frac{2}{k} ( e ^{- t} + \frac{k}{2} ) - 1}{2}))

Simplify

- 4 (- \frac{1}{2 k} (\frac{ln \frac{2}{k} ( e ^{- t} + \frac{k}{2} ) + 1}{2} - \frac{ln \frac{2}{k} ( e ^{- t} + \frac{k}{2} ) - 1}{2})) : \frac{1}{k} (ln e^{- t} \frac{2}{k} + 2 - ln (\frac{2}{∣ k ∣}) + t)

= \frac{1}{k} (ln e^{- t} \frac{2}{k} + 2 - ln (\frac{2}{∣ k ∣}) + t)

Add a constant to the solution

= \frac{1}{k} (ln e^{- t} \frac{2}{k} + 2 - ln (\frac{2}{∣ k ∣}) + t) + C

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