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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{2 ( x - 1 )} d x = \frac{1}{2} ln ∣ x - 1 ∣

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

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

Add a constant to the solution

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

\int (x + 1)^{3} (x - 2) d x

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\int \frac{( 3 x ^{4} - 3 x ^{2} + x )}{x} d x

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

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