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

The integral of (1-x)/(x^2-x+1) is -1/2 ln|4x^2-4x+4|+1/(sqrt(3))arctan(1/(sqrt(3))(2x-1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int x \frac{1}{1 + x ^{2}} d x

\int \frac{3 x + 7}{( x + 5 ) ( x + 1 )} d x

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

\int 2 e^{6 x} d x

\int x^{2} (ln (x) + 1) d x

What is the integral of (1-x)/(x^2-x+1) ?
The integral of (1-x)/(x^2-x+1) is -1/2 ln|4x^2-4x+4|+1/(sqrt(3))arctan(1/(sqrt(3))(2x-1))+C