What is the integral of (30)/(x^3-x^2-6x) ?

The integral of (30)/(x^3-x^2-6x) is 30(-1/6 ln|x|+1/10 ln|x+2|+1/15 ln|x-3|)+C

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integral of (30)/(x^3-x^2-6x)

\int \frac{30}{x ^{3} - x ^{2} - 6 x} d x

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

Solution steps

\int \frac{30}{x ^{3} - x ^{2} - 6 x} d x

Take the constant out:

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

= 30 \cdot \int \frac{1}{x ^{3} - x ^{2} - 6 x} d x

Take the partial fraction of

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

= 30 \cdot \int - \frac{1}{6 x} + \frac{1}{10 ( x + 2 )} + \frac{1}{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

= 30 (- \int \frac{1}{6 x} d x + \int \frac{1}{10 ( x + 2 )} d x + \int \frac{1}{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{1}{15 ( x - 3 )} d x = \frac{1}{15} ln ∣ x - 3 ∣

= 30 (- \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣)

Add a constant to the solution

= 30 (- \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣) + C

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What is the integral of (30)/(x^3-x^2-6x) ?
The integral of (30)/(x^3-x^2-6x) is 30(-1/6 ln|x|+1/10 ln|x+2|+1/15 ln|x-3|)+C