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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Expand

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

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

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

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

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

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

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

\int \frac{30}{2 x + 4} d x

\int x (2 x - 5)^{5} d x

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

\int \frac{3}{2 + x ^{2}} d x

\int \frac{5}{2 x ^{2} - 4 x + 6} d x

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