What is the integral of (2x-5)/(x^2+x+1) ?

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

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integral of (2x-5)/(x^2+x+1)

\int \frac{2 x - 5}{x ^{2} + x + 1} d x

ln 4 x^{2} + 4 x + 4 - 43 arctan (\frac{1}{3} (2 x + 1)) + C

Solution steps

\int \frac{2 x - 5}{x ^{2} + x + 1} d x

Expand

\frac{2 x - 5}{x ^{2} + x + 1} : \frac{2 x}{x ^{2} + x + 1} - \frac{5}{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{2 x}{x ^{2} + x + 1} d x - \int \frac{5}{x ^{2} + x + 1} d x

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

\int \frac{5}{x ^{2} + x + 1} d x = \frac{10}{3} arctan (\frac{2 x + 1}{3})

= ln 4 x^{2} + 4 x + 4 - \frac{2}{3} arctan (\frac{2 x + 1}{3}) - \frac{10}{3} arctan (\frac{2 x + 1}{3})

Simplify

ln 4 x^{2} + 4 x + 4 - \frac{2}{3} arctan (\frac{2 x + 1}{3}) - \frac{10}{3} arctan (\frac{2 x + 1}{3}) : ln 4 x^{2} + 4 x + 4 - 43 arctan (\frac{1}{3} (2 x + 1))

= ln 4 x^{2} + 4 x + 4 - 43 arctan (\frac{1}{3} (2 x + 1))

Add a constant to the solution

= ln 4 x^{2} + 4 x + 4 - 43 arctan (\frac{1}{3} (2 x + 1)) + C

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