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

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

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

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

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

Solution steps

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

Expand

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

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

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

= \frac{4}{3} ln x^{3} - 1 + \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

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

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

Add a constant to the solution

= \frac{4}{3} ln x^{3} - 1 + \frac{5}{3} 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}{6} (- ln \frac{4 x ^{2} + 4 x}{3} + \frac{4}{3} + 23 arctan (\frac{2 x + 1}{3})) + C

y = (1 + 4 x)^{10}, (0, 1)

3 \frac{d y}{d x} = e^{y - 6 x}

\frac{d}{d x} (2 (2 x e^{2 x} + e^{2 x} \cdot 2 x^{2}))

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

\int x sec (x^{2}) tan (x^{2}) d x

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