What is the integral of (3x^2-4x-1)/((x^2-1)(x-2)) ?

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

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integral of (3x^2-4x-1)/((x^2-1)(x-2))

\int \frac{3 x ^{2} - 4 x - 1}{( x ^{2} - 1 ) ( x - 2 )} d x

ln (∣ x - 2 ∣ x^{2} - 1) + C

Solution steps

\int \frac{3 x ^{2} - 4 x - 1}{( x ^{2} - 1 ) ( x - 2 )} d x

Apply u-substitution

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = (x^{2} - 1) (x - 2)

= ln (x^{2} - 1) (x - 2)

Simplify

ln (x^{2} - 1) (x - 2) : ln (∣ x - 2 ∣ x^{2} - 1)

= ln (∣ x - 2 ∣ x^{2} - 1)

Add a constant to the solution

= ln (∣ x - 2 ∣ x^{2} - 1) + C

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