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

The integral of 6/(x^3+1) is 2ln|x+1|-ln| 4/3+1/3 (4x^2-4x)|+2sqrt(3)arctan(1/(sqrt(3))(2x-1))+C

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integral of 6/(x^3+1)

\int \frac{6}{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)) + C

Solution steps

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

Take the constant out:

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

= 6 \cdot \int \frac{1}{x ^{3} + 1} d x

Take the partial fraction of

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

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

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

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

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

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

Simplify

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

Add a constant to the solution

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

What is the integral of 6/(x^3+1) ?
The integral of 6/(x^3+1) is 2ln|x+1|-ln| 4/3+1/3 (4x^2-4x)|+2sqrt(3)arctan(1/(sqrt(3))(2x-1))+C