What is the integral of 1/((x^2-1)(x-3)) ?

The integral of 1/((x^2-1)(x-3)) is 1/8 ln|x-3|-1/4 ln|x-1|+1/8 ln|x+1|+C

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integral of 1/((x^2-1)(x-3))

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

\frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{8} ln ∣ x + 1 ∣ + C

Solution steps

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

Apply u-substitution

= \int \frac{1}{u ( u ^{2} + 6 u + 8 )} d u

Take the partial fraction of

\frac{1}{u ( u ^{2} + 6 u + 8 )} : \frac{1}{8 u} - \frac{1}{4 ( u + 2 )} + \frac{1}{8 ( u + 4 )}

= \int \frac{1}{8 u} - \frac{1}{4 ( u + 2 )} + \frac{1}{8 ( u + 4 )} d u

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}{8 u} d u - \int \frac{1}{4 ( u + 2 )} d u + \int \frac{1}{8 ( u + 4 )} d u

\int \frac{1}{8 u} d u = \frac{1}{8} ln ∣ u ∣

\int \frac{1}{4 ( u + 2 )} d u = \frac{1}{4} ln ∣ u + 2 ∣

\int \frac{1}{8 ( u + 4 )} d u = \frac{1}{8} ln ∣ u + 4 ∣

= \frac{1}{8} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 2 ∣ + \frac{1}{8} ln ∣ u + 4 ∣

Substitute back

u = x - 3

= \frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 3 + 2 ∣ + \frac{1}{8} ln ∣ x - 3 + 4 ∣

Simplify

\frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 3 + 2 ∣ + \frac{1}{8} ln ∣ x - 3 + 4 ∣ : \frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{8} ln ∣ x + 1 ∣

= \frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{8} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ x - 3 ∣ - \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{8} ln ∣ x + 1 ∣ + C

What is the integral of 1/((x^2-1)(x-3)) ?
The integral of 1/((x^2-1)(x-3)) is 1/8 ln|x-3|-1/4 ln|x-1|+1/8 ln|x+1|+C