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

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

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integral of (ae^θ+b)/(ae^θ-b)

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

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

Solution steps

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

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

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

Take the constant out:

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

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

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 = 1 + \frac{a e ^{θ}}{b}

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

Simplify

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

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

Add a constant to the solution

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

\int 13 d x

\frac{1}{2 + 7 x}

\frac{d}{d x} (x^{3} ln ∣ 1 + x ∣)

\int \frac{x ^{2} - 4 x + 9}{x ^{2} - 6 x + 9} d x

arctan (9 x)

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