What is the integral of (x^5-x^4+4x^2-x)/(x^3+1) ?

The integral of (x^5-x^4+4x^2-x)/(x^3+1) is

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integral of (x^5-x^4+4x^2-x)/(x^3+1)

\int \frac{x ^{5} - x ^{4} + 4 x ^{2} - x}{x ^{3} + 1} d x

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

Solution steps

\int \frac{x ^{5} - x ^{4} + 4 x ^{2} - x}{x ^{3} + 1} d x

Expand

\frac{x ^{5} - x ^{4} + 4 x ^{2} - x}{x ^{3} + 1} : \frac{x ^{5}}{x ^{3} + 1} - \frac{x ^{4}}{x ^{3} + 1} + \frac{4 x ^{2}}{x ^{3} + 1} - \frac{x}{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{x ^{5}}{x ^{3} + 1} d x - \int \frac{x ^{4}}{x ^{3} + 1} d x + \int \frac{4 x ^{2}}{x ^{3} + 1} d x - \int \frac{x}{x ^{3} + 1} d x

\int \frac{x ^{5}}{x ^{3} + 1} d x = \frac{1}{3} (x^{3} + 1 - ln x^{3} + 1)

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

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

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (x cos (\frac{7}{x}))

\frac{x + x}{x ^{2}}

\frac{d}{d x} (ln (4 + x))

\int x e^{x} - e^{x} d x

\frac{d y}{d x} = 2 x y + \frac{6}{π}

What is the integral of (x^5-x^4+4x^2-x)/(x^3+1) ?
The integral of (x^5-x^4+4x^2-x)/(x^3+1) is