What is the integral of 1/(a+e^{ax)} ?

The integral of 1/(a+e^{ax)} is 1/(a^2)(-ln((|a+e^{ax}|}{|a|})+ln|\frac{a+e^{ax})/a-1|)+C

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integral of 1/(a+e^{ax)}

\int \frac{1}{a + e ^{a x}} d x

\frac{1}{a ^{2}} (- ln (\frac{∣ a + e ^{a x} ∣}{∣ a ∣}) + ln \frac{a + e ^{a x}}{a} - 1) + C

Solution steps

\int \frac{1}{a + e ^{a x}} d x

Apply u-substitution

: \int \frac{1}{a ( a + e ^{u} )} d u

= \int \frac{1}{a ( a + e ^{u} )} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int \frac{1}{a + e ^{u}} d u

Apply u-substitution

: \int \frac{1}{v ( v - a )} d v

= \frac{1}{a} \cdot \int \frac{1}{v ( v - a )} d v

v - a = a (\frac{v}{a} - 1)

= \frac{1}{a} \cdot \int \frac{1}{v a ( \frac{v}{a} - 1 )} d v

Take the constant out:

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

= \frac{1}{a} \cdot \frac{1}{a} \cdot \int \frac{1}{v ( \frac{v}{a} - 1 )} d v

Apply u-substitution

: \int \frac{1}{w ( w - 1 )} d w

= \frac{1}{a} \cdot \frac{1}{a} \cdot \int \frac{1}{w ( w - 1 )} d w

Take the partial fraction of

\frac{1}{w ( w - 1 )} : - \frac{1}{w} + \frac{1}{w - 1}

= \frac{1}{a} \cdot \frac{1}{a} \cdot \int - \frac{1}{w} + \frac{1}{w - 1} d w

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}{a} \cdot \frac{1}{a} (- \int \frac{1}{w} d w + \int \frac{1}{w - 1} d w)

\int \frac{1}{w} d w = ln ∣ w ∣

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

= \frac{1}{a} \cdot \frac{1}{a} (- ln ∣ w ∣ + ln ∣ w - 1 ∣)

Substitute back

= \frac{1}{a} \cdot \frac{1}{a} (- ln \frac{a + e ^{a x}}{a} + ln \frac{a + e ^{a x}}{a} - 1)

Simplify

\frac{1}{a} \cdot \frac{1}{a} (- ln \frac{a + e ^{a x}}{a} + ln \frac{a + e ^{a x}}{a} - 1) : \frac{1}{a ^{2}} (- ln (\frac{∣ a + e ^{a x} ∣}{∣ a ∣}) + ln \frac{a + e ^{a x}}{a} - 1)

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

Add a constant to the solution

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

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What is the integral of 1/(a+e^{ax)} ?
The integral of 1/(a+e^{ax)} is 1/(a^2)(-ln((|a+e^{ax}|}{|a|})+ln|\frac{a+e^{ax})/a-1|)+C