What is the integral of 1/(v(v+b)) ?

The integral of 1/(v(v+b)) is 1/b (-ln((|b+v|}{|b|})+ln|\frac{b+v)/b-1|)+C

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integral of 1/(v(v+b))

\int \frac{1}{v ( v + b )} d v

\frac{1}{b} (- ln (\frac{∣ b + v ∣}{∣ b ∣}) + ln \frac{b + v}{b} - 1) + C

Solution steps

\int \frac{1}{v ( v + b )} d v

Apply u-substitution

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

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

u - b = b (\frac{u}{b} - 1)

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

Take the constant out:

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

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

Apply u-substitution

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

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

Substitute back

= \frac{1}{b} (- ln \frac{b + v}{b} + ln \frac{b + v}{b} - 1)

Apply absolute rule:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{b + v}{b} = \frac{∣ b + v ∣}{∣ b ∣}

= \frac{1}{b} (- ln (\frac{∣ b + v ∣}{∣ b ∣}) + ln \frac{b + v}{b} - 1)

Add a constant to the solution

= \frac{1}{b} (- ln (\frac{∣ b + v ∣}{∣ b ∣}) + ln \frac{b + v}{b} - 1) + C

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What is the integral of 1/(v(v+b)) ?
The integral of 1/(v(v+b)) is 1/b (-ln((|b+v|}{|b|})+ln|\frac{b+v)/b-1|)+C