What is the integral of v/(1-kv) ?

The integral of v/(1-kv) is 1/(k^2)(1-kv-ln|1-kv|)+C

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integral of v/(1-kv)

\int \frac{v}{1 - k v} d v

\frac{1}{k ^{2}} (1 - k v - ln ∣ 1 - k v ∣) + C

Solution steps

\int \frac{v}{1 - k v} d v

Apply u-substitution

: \int \frac{u - 1}{k ^{2} u} d u

= \int \frac{u - 1}{k ^{2} u} d u

Take the constant out:

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

= \frac{1}{k ^{2}} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{1}{k ^{2}} \cdot \int 1 - \frac{1}{u} 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

= \frac{1}{k ^{2}} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{k ^{2}} (u - ln ∣ u ∣)

Substitute back

u = 1 - k v

= \frac{1}{k ^{2}} (1 - k v - ln ∣ 1 - k v ∣)

Add a constant to the solution

= \frac{1}{k ^{2}} (1 - k v - ln ∣ 1 - k v ∣) + C

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What is the integral of v/(1-kv) ?
The integral of v/(1-kv) is 1/(k^2)(1-kv-ln|1-kv|)+C