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

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

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

Take the partial fraction of

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

\int \frac{2 x}{5 x + 3} d x

\int e^{a x} sin (x) d x

\int a^{3 x} d x

\int \frac{1}{3 x ^{2} + 5 x + 1} d x

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