What is the integral of (2x+5)/(x^3-x^2-12x) ?

The integral of (2x+5)/(x^3-x^2-12x) is -1/21 ln|x+3|+13/28 ln|x-4|-5/12 ln|x|+C

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integral of (2x+5)/(x^3-x^2-12x)

\int \frac{2 x + 5}{x ^{3} - x ^{2} - 12 x} d x

- \frac{1}{21} ln ∣ x + 3 ∣ + \frac{13}{28} ln ∣ x - 4 ∣ - \frac{5}{12} ln ∣ x ∣ + C

Solution steps

\int \frac{2 x + 5}{x ^{3} - x ^{2} - 12 x} d x

Expand

\frac{2 x + 5}{x ^{3} - x ^{2} - 12 x} : \frac{2 x}{x ^{3} - x ^{2} - 12 x} + \frac{5}{x ^{3} - x ^{2} - 12 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{2 x}{x ^{3} - x ^{2} - 12 x} d x + \int \frac{5}{x ^{3} - x ^{2} - 12 x} d x

\int \frac{2 x}{x ^{3} - x ^{2} - 12 x} d x = 2 (- \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{7} ln ∣ x - 4 ∣)

\int \frac{5}{x ^{3} - x ^{2} - 12 x} d x = 5 (- \frac{1}{12} ln ∣ x ∣ + \frac{1}{21} ln ∣ x + 3 ∣ + \frac{1}{28} ln ∣ x - 4 ∣)

= 2 (- \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{7} ln ∣ x - 4 ∣) + 5 (- \frac{1}{12} ln ∣ x ∣ + \frac{1}{21} ln ∣ x + 3 ∣ + \frac{1}{28} ln ∣ x - 4 ∣)

Simplify

2 (- \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{7} ln ∣ x - 4 ∣) + 5 (- \frac{1}{12} ln ∣ x ∣ + \frac{1}{21} ln ∣ x + 3 ∣ + \frac{1}{28} ln ∣ x - 4 ∣) : - \frac{1}{21} ln ∣ x + 3 ∣ + \frac{13}{28} ln ∣ x - 4 ∣ - \frac{5}{12} ln ∣ x ∣

= - \frac{1}{21} ln ∣ x + 3 ∣ + \frac{13}{28} ln ∣ x - 4 ∣ - \frac{5}{12} ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{21} ln ∣ x + 3 ∣ + \frac{13}{28} ln ∣ x - 4 ∣ - \frac{5}{12} ln ∣ x ∣ + C

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What is the integral of (2x+5)/(x^3-x^2-12x) ?
The integral of (2x+5)/(x^3-x^2-12x) is -1/21 ln|x+3|+13/28 ln|x-4|-5/12 ln|x|+C