What is the integral of 1/(kx-nx^2) ?

The integral of 1/(kx-nx^2) is 1/k (ln((|n||x|}{|k|})-ln|\frac{nx)/k-1|)+C

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integral of 1/(kx-nx^2)

\int \frac{1}{k x - n x ^{2}} d x

\frac{1}{k} (ln (\frac{∣ n ∣ ∣ x ∣}{∣ k ∣}) - ln \frac{n x}{k} - 1) + C

Solution steps

\int \frac{1}{k x - n x ^{2}} d x

Factor

\frac{1}{k x - n x ^{2}}

= \int \frac{1}{x ( k - n x )} d x

k - n x = k (1 - \frac{n x}{k})

= \int \frac{1}{x k ( 1 - \frac{n x}{k} )} d x

Take the constant out:

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

= \frac{1}{k} \cdot \int \frac{1}{x ( 1 - \frac{n x}{k} )} d x

Apply u-substitution

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

Take the partial fraction of

\frac{1}{u ( - u + 1 )} : \frac{1}{u} - \frac{1}{u - 1}

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

Substitute back

u = \frac{n x}{k}

= \frac{1}{k} (ln \frac{n x}{k} - ln \frac{n x}{k} - 1)

Simplify

\frac{n x}{k} : \frac{∣ n ∣ ∣ x ∣}{∣ k ∣}

= \frac{1}{k} (ln (\frac{∣ n ∣ ∣ x ∣}{∣ k ∣}) - ln \frac{n x}{k} - 1)

Add a constant to the solution

= \frac{1}{k} (ln (\frac{∣ n ∣ ∣ x ∣}{∣ k ∣}) - ln \frac{n x}{k} - 1) + C

What is the integral of 1/(kx-nx^2) ?
The integral of 1/(kx-nx^2) is 1/k (ln((|n||x|}{|k|})-ln|\frac{nx)/k-1|)+C