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

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

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

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

ln ∣ 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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\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))

= ln ∣ x ∣ - \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

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

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