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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

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

Add a constant to the solution

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

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