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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

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

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

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

Add a constant to the solution

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

2

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