What is the integral of (x^2+x+6)/(x^3-8) ?

The integral of (x^2+x+6)/(x^3-8) is

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integral of (x^2+x+6)/(x^3-8)

\int \frac{x ^{2} + x + 6}{x ^{3} - 8} d x

\frac{1}{3} ln x^{3} - 8 + \frac{2}{3} ln ∣ x - 2 ∣ - \frac{1}{3} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C

Solution steps

\int \frac{x ^{2} + x + 6}{x ^{3} - 8} d x

Expand

\frac{x ^{2} + x + 6}{x ^{3} - 8} : \frac{x ^{2}}{x ^{3} - 8} + \frac{x}{x ^{3} - 8} + \frac{6}{x ^{3} - 8}

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{x ^{2}}{x ^{3} - 8} d x + \int \frac{x}{x ^{3} - 8} d x + \int \frac{6}{x ^{3} - 8} d x

\int \frac{x ^{2}}{x ^{3} - 8} d x = \frac{1}{3} ln x^{3} - 8

\int \frac{x}{x ^{3} - 8} d x = \frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{12} ln x^{2} + 2 x + 4 + \frac{1}{2 3} arctan (\frac{1}{3} (x + 1))

\int \frac{6}{x ^{3} - 8} d x = 6 (\frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1)))

= \frac{1}{3} ln x^{3} - 8 + \frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{12} ln x^{2} + 2 x + 4 + \frac{1}{2 3} arctan (\frac{1}{3} (x + 1)) + 6 (\frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1)))

Simplify

\frac{1}{3} ln x^{3} - 8 + \frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{12} ln x^{2} + 2 x + 4 + \frac{1}{2 3} arctan (\frac{1}{3} (x + 1)) + 6 (\frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1))) : \frac{1}{3} ln x^{3} - 8 + \frac{2}{3} ln ∣ x - 2 ∣ - \frac{1}{3} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1))

= \frac{1}{3} ln x^{3} - 8 + \frac{2}{3} ln ∣ x - 2 ∣ - \frac{1}{3} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1))

Add a constant to the solution

= \frac{1}{3} ln x^{3} - 8 + \frac{2}{3} ln ∣ x - 2 ∣ - \frac{1}{3} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C

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What is the integral of (x^2+x+6)/(x^3-8) ?
The integral of (x^2+x+6)/(x^3-8) is