What is the integral of (9x^2+6x+9)/(x^3-1) ?

The integral of (9x^2+6x+9)/(x^3-1) is

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integral of (9x^2+6x+9)/(x^3-1)

\int \frac{9 x ^{2} + 6 x + 9}{x ^{3} - 1} d x

- \frac{3}{2} ln 4 x^{2} + 4 x + 4 + 3 ln x^{3} - 1 + 5 ln ∣ x - 1 ∣ - ln \frac{4}{3} + \frac{1}{3} (4 x^{2} + 4 x) - 3 arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

\int \frac{9 x ^{2} + 6 x + 9}{x ^{3} - 1} d x

Expand

\frac{9 x ^{2} + 6 x + 9}{x ^{3} - 1} : \frac{9 x ^{2}}{x ^{3} - 1} + \frac{6 x}{x ^{3} - 1} + \frac{9}{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{9 x ^{2}}{x ^{3} - 1} d x + \int \frac{6 x}{x ^{3} - 1} d x + \int \frac{9}{x ^{3} - 1} d x

\int \frac{9 x ^{2}}{x ^{3} - 1} d x = 3 ln x^{3} - 1

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

\int \frac{9}{x ^{3} - 1} d x = 9 (\frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{6} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1)))

= 3 ln x^{3} - 1 + 2 ln ∣ x - 1 ∣ - ln \frac{4}{3} + \frac{1}{3} (4 x^{2} + 4 x) + 23 arctan (\frac{1}{3} (2 x + 1)) + 9 (\frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{6} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1)))

Simplify

3 ln x^{3} - 1 + 2 ln ∣ x - 1 ∣ - ln \frac{4}{3} + \frac{1}{3} (4 x^{2} + 4 x) + 23 arctan (\frac{1}{3} (2 x + 1)) + 9 (\frac{1}{3} ln ∣ x - 1 ∣ - \frac{1}{6} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 x + 1))) : - \frac{3}{2} ln 4 x^{2} + 4 x + 4 + 3 ln x^{3} - 1 + 5 ln ∣ x - 1 ∣ - ln \frac{4}{3} + \frac{1}{3} (4 x^{2} + 4 x) - 3 arctan (\frac{1}{3} (2 x + 1))

= - \frac{3}{2} ln 4 x^{2} + 4 x + 4 + 3 ln x^{3} - 1 + 5 ln ∣ x - 1 ∣ - ln \frac{4}{3} + \frac{1}{3} (4 x^{2} + 4 x) - 3 arctan (\frac{1}{3} (2 x + 1))

Add a constant to the solution

= - \frac{3}{2} ln 4 x^{2} + 4 x + 4 + 3 ln x^{3} - 1 + 5 ln ∣ x - 1 ∣ - ln \frac{4}{3} + \frac{1}{3} (4 x^{2} + 4 x) - 3 arctan (\frac{1}{3} (2 x + 1)) + C

What is the integral of (9x^2+6x+9)/(x^3-1) ?
The integral of (9x^2+6x+9)/(x^3-1) is