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

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

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

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

Expand

\frac{- x - 1}{x ^{2} + x + 1} : - \frac{x}{x ^{2} + x + 1} - \frac{1}{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{1}{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{1}{x ^{2} + x + 1} d x = \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{2}{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{2}{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^{5} cos (2 x) d x

y = (1 + 4 x)^{\frac{3}{2}}, at x = 2

f (x) = 5 - 8 x^{2}, (- 2, - 27)

(x + 2 y) d x + (2 x - y) d y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0

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