What is the integral of (x-1)/(x^3+x^2-6x) ?

The integral of (x-1)/(x^3+x^2-6x) is 1/6 ln|x|+1/10 ln|x-2|-4/15 ln|x+3|+C

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integral of (x-1)/(x^3+x^2-6x)

\int \frac{x - 1}{x ^{3} + x ^{2} - 6 x} d x

\frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x - 2 ∣ - \frac{4}{15} ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{x - 1}{x ^{3} + x ^{2} - 6 x} d x

Take the partial fraction of

\frac{x - 1}{x ^{3} + x ^{2} - 6 x} : \frac{1}{6 x} + \frac{1}{10 ( x - 2 )} - \frac{4}{15 ( x + 3 )}

= \int \frac{1}{6 x} + \frac{1}{10 ( x - 2 )} - \frac{4}{15 ( x + 3 )} 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}{6 x} d x + \int \frac{1}{10 ( x - 2 )} d x - \int \frac{4}{15 ( x + 3 )} d x

\int \frac{1}{6 x} d x = \frac{1}{6} ln ∣ x ∣

\int \frac{1}{10 ( x - 2 )} d x = \frac{1}{10} ln ∣ x - 2 ∣

\int \frac{4}{15 ( x + 3 )} d x = \frac{4}{15} ln ∣ x + 3 ∣

= \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x - 2 ∣ - \frac{4}{15} ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x - 2 ∣ - \frac{4}{15} ln ∣ x + 3 ∣ + C

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What is the integral of (x-1)/(x^3+x^2-6x) ?
The integral of (x-1)/(x^3+x^2-6x) is 1/6 ln|x|+1/10 ln|x-2|-4/15 ln|x+3|+C