What is the integral of (2x^4-x-4)/(x^3+8) ?

The integral of (2x^4-x-4)/(x^3+8) is

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integral of (2x^4-x-4)/(x^3+8)

\int \frac{2 x ^{4} - x - 4}{x ^{3} + 8} d x

x^{2} + \frac{5}{2} ln ∣ x + 2 ∣ - \frac{17}{12} ln x^{2} - 2 x + 4 + \frac{1}{6} ln \frac{1}{3} (x - 1)^{2} + 1 - \frac{19}{2 3} arctan (\frac{1}{3} (x - 1)) + C

Solution steps

\int \frac{2 x ^{4} - x - 4}{x ^{3} + 8} d x

Expand

\frac{2 x ^{4} - x - 4}{x ^{3} + 8} : \frac{2 x ^{4}}{x ^{3} + 8} - \frac{x}{x ^{3} + 8} - \frac{4}{x ^{3} + 8}

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 ^{4}}{x ^{3} + 8} d x - \int \frac{x}{x ^{3} + 8} d x - \int \frac{4}{x ^{3} + 8} d x

\int \frac{2 x ^{4}}{x ^{3} + 8} d x = x^{2} + \frac{8}{3} ln ∣ x + 2 ∣ - \frac{4}{3} ln x^{2} - 2 x + 4 - \frac{8}{3} arctan (\frac{1}{3} (x - 1))

\int \frac{x}{x ^{3} + 8} d x = - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{12} (ln x^{2} - 2 x + 4 + 23 arctan (\frac{1}{3} (x - 1)))

\int \frac{4}{x ^{3} + 8} d x = 4 (\frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} (- \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1))))

= x^{2} + \frac{8}{3} ln ∣ x + 2 ∣ - \frac{4}{3} ln x^{2} - 2 x + 4 - \frac{8}{3} arctan (\frac{1}{3} (x - 1)) - (- \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{12} (ln x^{2} - 2 x + 4 + 23 arctan (\frac{1}{3} (x - 1)))) - 4 (\frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} (- \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1))))

Simplify

x^{2} + \frac{8}{3} ln ∣ x + 2 ∣ - \frac{4}{3} ln x^{2} - 2 x + 4 - \frac{8}{3} arctan (\frac{1}{3} (x - 1)) - (- \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{12} (ln x^{2} - 2 x + 4 + 23 arctan (\frac{1}{3} (x - 1)))) - 4 (\frac{1}{12} ln ∣ x + 2 ∣ + \frac{1}{12} (- \frac{1}{2} ln \frac{1}{3} (x - 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x - 1)))) : x^{2} + \frac{5}{2} ln ∣ x + 2 ∣ - \frac{17}{12} ln x^{2} - 2 x + 4 + \frac{1}{6} ln \frac{1}{3} (x - 1)^{2} + 1 - \frac{19}{2 3} arctan (\frac{1}{3} (x - 1))

= x^{2} + \frac{5}{2} ln ∣ x + 2 ∣ - \frac{17}{12} ln x^{2} - 2 x + 4 + \frac{1}{6} ln \frac{1}{3} (x - 1)^{2} + 1 - \frac{19}{2 3} arctan (\frac{1}{3} (x - 1))

Add a constant to the solution

= x^{2} + \frac{5}{2} ln ∣ x + 2 ∣ - \frac{17}{12} ln x^{2} - 2 x + 4 + \frac{1}{6} ln \frac{1}{3} (x - 1)^{2} + 1 - \frac{19}{2 3} arctan (\frac{1}{3} (x - 1)) + C

\int \frac{e ^{12 t}}{4 e ^{6 t} - 1} d t

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

\int \frac{2}{3} e^{- 2 t} d t

\int x^{3} sin (x^{4} + 2) d x

What is the integral of (2x^4-x-4)/(x^3+8) ?
The integral of (2x^4-x-4)/(x^3+8) is