What is the integral of (3x)/(x^3+2x^2-5x-6) ?

The integral of (3x)/(x^3+2x^2-5x-6) is 3(1/6 ln|x+1|+2/15 ln|x-2|-3/10 ln|x+3|)+C

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integral of (3x)/(x^3+2x^2-5x-6)

\int \frac{3 x}{x ^{3} + 2 x ^{2} - 5 x - 6} d x

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

Solution steps

\int \frac{3 x}{x ^{3} + 2 x ^{2} - 5 x - 6} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{3} + 2 x ^{2} - 5 x - 6} d x

Take the partial fraction of

\frac{x}{x ^{3} + 2 x ^{2} - 5 x - 6} : \frac{1}{6 ( x + 1 )} + \frac{2}{15 ( x - 2 )} - \frac{3}{10 ( x + 3 )}

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

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

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

\int \frac{2}{15 ( x - 2 )} d x = \frac{2}{15} ln ∣ x - 2 ∣

\int \frac{3}{10 ( x + 3 )} d x = \frac{3}{10} ln ∣ x + 3 ∣

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

Add a constant to the solution

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

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What is the integral of (3x)/(x^3+2x^2-5x-6) ?
The integral of (3x)/(x^3+2x^2-5x-6) is 3(1/6 ln|x+1|+2/15 ln|x-2|-3/10 ln|x+3|)+C