What is the integral of 1/((x^2-x)(x^2+4)) ?

The integral of 1/((x^2-x)(x^2+4)) is -1/4 ln|x|+1/5 ln|x-1|+1/20 (1/2 ln|(x^2)/4+1|-2arctan(x/2))+C

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integral of 1/((x^2-x)(x^2+4))

\int \frac{1}{( x ^{2} - x ) ( x ^{2} + 4 )} d x

- \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{20} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - 2 arctan (\frac{x}{2})) + C

Solution steps

\int \frac{1}{( x ^{2} - x ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{1}{( x ^{2} - x ) ( x ^{2} + 4 )} : - \frac{1}{4 x} + \frac{1}{5 ( x - 1 )} + \frac{x - 4}{20 ( x ^{2} + 4 )}

= \int - \frac{1}{4 x} + \frac{1}{5 ( x - 1 )} + \frac{x - 4}{20 ( x ^{2} + 4 )} 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}{4 x} d x + \int \frac{1}{5 ( x - 1 )} d x + \int \frac{x - 4}{20 ( x ^{2} + 4 )} d x

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{1}{5 ( x - 1 )} d x = \frac{1}{5} ln ∣ x - 1 ∣

\int \frac{x - 4}{20 ( x ^{2} + 4 )} d x = \frac{1}{20} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - 2 arctan (\frac{x}{2}))

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{20} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - 2 arctan (\frac{x}{2}))

Add a constant to the solution

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{20} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 - 2 arctan (\frac{x}{2})) + C

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What is the integral of 1/((x^2-x)(x^2+4)) ?
The integral of 1/((x^2-x)(x^2+4)) is -1/4 ln|x|+1/5 ln|x-1|+1/20 (1/2 ln|(x^2)/4+1|-2arctan(x/2))+C