What is the integral of 1/(2x^2-x^3) ?

The integral of 1/(2x^2-x^3) is -1/4 (-ln|x|+2/x)-1/4 ln|x-2|+C

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integral of 1/(2x^2-x^3)

\int \frac{1}{2 x ^{2} - x ^{3}} d x

- \frac{1}{4} (- ln ∣ x ∣ + \frac{2}{x}) - \frac{1}{4} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{1}{2 x ^{2} - x ^{3}} d x

Take the partial fraction of

\frac{1}{2 x ^{2} - x ^{3}} : - \frac{- x - 2}{4 x ^{2}} - \frac{1}{4 ( x - 2 )}

= \int - \frac{- x - 2}{4 x ^{2}} - \frac{1}{4 ( x - 2 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{- x - 2}{4 x ^{2}} d x - \int \frac{1}{4 ( x - 2 )} d x

\int \frac{- x - 2}{4 x ^{2}} d x = \frac{1}{4} (- ln ∣ x ∣ + \frac{2}{x})

\int \frac{1}{4 ( x - 2 )} d x = \frac{1}{4} ln ∣ x - 2 ∣

= - \frac{1}{4} (- ln ∣ x ∣ + \frac{2}{x}) - \frac{1}{4} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{1}{4} (- ln ∣ x ∣ + \frac{2}{x}) - \frac{1}{4} ln ∣ x - 2 ∣ + C

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What is the integral of 1/(2x^2-x^3) ?
The integral of 1/(2x^2-x^3) is -1/4 (-ln|x|+2/x)-1/4 ln|x-2|+C