Integrate dx/(x-x^2/3)

\(\int \frac{x-x^2}{3}\)

First, remove the denominator outside the integral, it won't affect the outcome:

\(\frac{1}{3}\left(\int x-x^2\right)\)

Integrate each term individually, you're allowed to do that:

\(\frac{1}{3}\left(\int x-\int x^2\right)\)

The integral of both of these is easy, \(x\) becomes \(\frac{x^2}{2}\) and \(x^2\) becomes \(\frac{x^3}{3}\)

Therefore:

\(\frac{1}{3}\left(\frac{x^2}{2}-\frac{x^3}{3}\right)\)

Multiply the 1/3 back into the parentheses:

\(\frac{x^2}{6}-\frac{x^3}{9}\)

You can find a common denominator between 6 and 9, it's 18:

\(\frac{x^2}{6}-\frac{x^3}{9}\rightarrow \ \frac{9x^2-6x^3}{18}\)

Common factor the denominator:

\(\frac{x^2}{6}-\frac{x^3}{9}\rightarrow \ \frac{9x^2-6x^3}{18}\rightarrow \ \frac{3x^2\left(3-2x\right)}{18}\)

Cancel out the 3 and 18:

\(\frac{x^2\left(3-2x\right)}{6}+C\)