What is the integral of (e^{ax}+b)/(e^{ax)-b} ?

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

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integral of (e^{ax}+b)/(e^{ax)-b}

\int \frac{e ^{a x} + b}{e ^{a x} - b} d x

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

Solution steps

\int \frac{e ^{a x} + b}{e ^{a x} - b} d x

e^{a x} + b = b (\frac{e ^{a x}}{b} + 1)

= \int \frac{b ( \frac{e ^{a x}}{b} + 1 )}{e ^{a x} - b} d x

Take the constant out:

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

= b \cdot \int \frac{\frac{e ^{a x}}{b} + 1}{e ^{a x} - b} d x

Apply u-substitution

: \int \frac{u}{ab ( u - 1 ) ( u - 2 )} d u

= b \cdot \int \frac{u}{ab ( u - 1 ) ( u - 2 )} d u

Take the constant out:

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

= b \frac{1}{ab} \cdot \int \frac{u}{( u - 1 ) ( u - 2 )} d u

Take the partial fraction of

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

= b \frac{1}{ab} \cdot \int - \frac{1}{u - 1} + \frac{2}{u - 2} d u

Apply the Sum Rule:

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

= b \frac{1}{ab} (- \int \frac{1}{u - 1} d u + \int \frac{2}{u - 2} d u)

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

\int \frac{2}{u - 2} d u = 2 ln ∣ u - 2 ∣

= b \frac{1}{ab} (- ln ∣ u - 1 ∣ + 2 ln ∣ u - 2 ∣)

Substitute back

u = \frac{e ^{a x}}{b} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int 10 x^{4} - 2 x - 6 d x

\int x x + 6 d x

\frac{d}{d x} (\frac{x ^{2}}{1 + x ^{3}})

(tan^{3} (x))^{^{'}}

4 t e^{- 2 t}

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