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

The integral of (x-2)/(x^2+x+1) is 1/2 ln|4x^2+4x+4|-5/(sqrt(3))arctan(1/(sqrt(3))(2x+1))+C

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

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

\frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

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

Expand

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

\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})

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

Simplify

\frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{2 x + 1}{3}) - \frac{4}{3} arctan (\frac{2 x + 1}{3}) : \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{3} arctan (\frac{1}{3} (2 x + 1))

= \frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{5}{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{5}{3} arctan (\frac{1}{3} (2 x + 1)) + C

\frac{d}{d x} (- csc^{2} (x) - cos (2 x) \cdot 2)

(y^{2} + 1) d x = y (sec^{2} (x)) d y

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

\int \frac{9 + x + x}{x} d x

n = 1 \sum \infty arctan (19 n)

What is the integral of (x-2)/(x^2+x+1) ?
The integral of (x-2)/(x^2+x+1) is 1/2 ln|4x^2+4x+4|-5/(sqrt(3))arctan(1/(sqrt(3))(2x+1))+C