What is the integral of (ae^x+b)/(ae^x-b) ?

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

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integral of (ae^x+b)/(ae^x-b)

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

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

Solution steps

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

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

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

Apply u-substitution

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

(9 x y^{2} - 4 y) + (9 x^{2} y - 4 x) y^{^{'}} = 0

\int (csc (3 x) - cot (3 x))^{2} d x

y^{^{''}} - 2 y^{^{'}} = sin (x), y (0) = 1, y^{^{'}} (0) = 2

\int t^{2} + 4 d t

\frac{d}{d x} (sin (x) + \frac{5}{6} cot (x))

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