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

The integral of x/((x-2)(x^2-1)) is 2/3 ln|x-2|-1/6 ln|x+1|-1/2 ln|x-1|+C

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

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

\frac{2}{3} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x}{( x - 2 ) ( x ^{2} - 1 )} : \frac{2}{3 ( x - 2 )} - \frac{1}{6 ( x + 1 )} - \frac{1}{2 ( x - 1 )}

= \int \frac{2}{3 ( x - 2 )} - \frac{1}{6 ( x + 1 )} - \frac{1}{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{2}{3 ( x - 2 )} d x - \int \frac{1}{6 ( x + 1 )} d x - \int \frac{1}{2 ( x - 1 )} d x

\int \frac{2}{3 ( x - 2 )} d x = \frac{2}{3} ln ∣ x - 2 ∣

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

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

= \frac{2}{3} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{2}{3} ln ∣ x - 2 ∣ - \frac{1}{6} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + C

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What is the integral of x/((x-2)(x^2-1)) ?
The integral of x/((x-2)(x^2-1)) is 2/3 ln|x-2|-1/6 ln|x+1|-1/2 ln|x-1|+C