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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Long division

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 4 d x = 4 x

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

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

Add a constant to the solution

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

\int e^{- 3 x^{2} - 1} x d x

\int \frac{3 x ^{6} + 4 x ^{2} + 12}{x ^{4}} d x

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

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

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

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