What is the integral of (3x^2-6x-4)/(x^3-3x^2-4x+12) ?

The integral of (3x^2-6x-4)/(x^3-3x^2-4x+12) is ln|x^3-3x^2-4x+12|+C

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integral of (3x^2-6x-4)/(x^3-3x^2-4x+12)

\int \frac{3 x ^{2} - 6 x - 4}{x ^{3} - 3 x ^{2} - 4 x + 12} d x

ln x^{3} - 3 x^{2} - 4 x + 12 + C

Solution steps

\int \frac{3 x ^{2} - 6 x - 4}{x ^{3} - 3 x ^{2} - 4 x + 12} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{3} - 3 x^{2} - 4 x + 12

= ln x^{3} - 3 x^{2} - 4 x + 12

Add a constant to the solution

= ln x^{3} - 3 x^{2} - 4 x + 12 + C

\frac{d}{d x} (x sinh (x))

x \to 2 + lim (\frac{x ^{3} - 8}{x - 2})

\int arccot (11 y) d y

f (x) = arctan (2 x)

\frac{d}{d x} ((x + 4) (e^{2} x))

What is the integral of (3x^2-6x-4)/(x^3-3x^2-4x+12) ?
The integral of (3x^2-6x-4)/(x^3-3x^2-4x+12) is ln|x^3-3x^2-4x+12|+C