What is the integral of ((2x+4))/((x^3-1)) ?

The integral of ((2x+4))/((x^3-1)) is

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integral of ((2x+4))/((x^3-1))

integral

\frac{( 2 x + 4 )}{( x ^{3} - 1 )}

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

Solution steps

\int \frac{2 x + 4}{x ^{3} - 1} d x

Expand

\frac{2 x + 4}{x ^{3} - 1} : \frac{2 x}{x ^{3} - 1} + \frac{4}{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{2 x}{x ^{3} - 1} d x + \int \frac{4}{x ^{3} - 1} d x

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

\int \frac{4}{x ^{3} - 1} d x = 4 (\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)))

= 2 (\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}))) + 4 (\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

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

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

Add a constant to the solution

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

(x^{2} + 5) \cdot 2^{x}

2 e^{- 2} x

3 x^{2} + 2 x + 1

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

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

What is the integral of ((2x+4))/((x^3-1)) ?
The integral of ((2x+4))/((x^3-1)) is