What is the integral of 1/(g-kv^2) ?

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

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integral of 1/(g-kv^2)

\int \frac{1}{g - k v ^{2}} d v

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

Solution steps

\int \frac{1}{g - k v ^{2}} d v

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{g k} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = \frac{k}{g} v

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

Simplify

\frac{1}{g k} \frac{ln \frac{k}{g} v + 1}{2} - \frac{ln \frac{k}{g} v - 1}{2} : \frac{1}{2 g k} (ln \frac{k}{g} v + 1 - ln \frac{k}{g} v - 1)

= \frac{1}{2 g k} (ln \frac{k}{g} v + 1 - ln \frac{k}{g} v - 1)

Add a constant to the solution

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

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