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

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

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

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

2 ln x^{3} - 1 + 3 ln ∣ x - 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} + 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x + 1}{3})) - ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

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

Expand

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

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

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

\int \frac{3 x}{x ^{3} - 1} d x = ln ∣ x - 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} + 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x + 1}{3}))

\int \frac{6}{x ^{3} - 1} d x = 2 ln ∣ x - 1 ∣ - ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1))

= 2 ln x^{3} - 1 + ln ∣ x - 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} + 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x + 1}{3})) + 2 ln ∣ x - 1 ∣ - ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1))

Simplify

2 ln x^{3} - 1 + ln ∣ x - 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} + 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x + 1}{3})) + 2 ln ∣ x - 1 ∣ - ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1)) : 2 ln x^{3} - 1 + 3 ln ∣ x - 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} + 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x + 1}{3})) - ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1))

= 2 ln x^{3} - 1 + 3 ln ∣ x - 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} + 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x + 1}{3})) - ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1))

Add a constant to the solution

= 2 ln x^{3} - 1 + 3 ln ∣ x - 1 ∣ + \frac{1}{2} (- ln \frac{4 x ^{2} + 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x + 1}{3})) - ln 4 x^{2} + 4 x + 4 - 23 arctan (\frac{1}{3} (2 x + 1)) + C

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