What is the integral of v/(g-kv^2) ?

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

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integral of v/(g-kv^2)

\int \frac{v}{g - k v ^{2}} d v

- \frac{1}{2 k} ln g - k v^{2} + C

Solution steps

\int \frac{v}{g - k v ^{2}} d v

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2 k} ln ∣ u ∣

Substitute back

u = g - k v^{2}

= - \frac{1}{2 k} ln g - k v^{2}

Add a constant to the solution

= - \frac{1}{2 k} ln g - k v^{2} + C

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\int (\frac{1}{x ^{2} - 1}) d x

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