What is the integral of (3x^2-2)/(x^3-2x-8) ?

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

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integral of (3x^2-2)/(x^3-2x-8)

\int \frac{3 x ^{2} - 2}{x ^{3} - 2 x - 8} d x

ln x^{3} - 2 x - 8 + C

Solution steps

\int \frac{3 x ^{2} - 2}{x ^{3} - 2 x - 8} 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^{3} - 2 x - 8

= ln x^{3} - 2 x - 8

Add a constant to the solution

= ln x^{3} - 2 x - 8 + C

x \to - \frac{π}{4} lim (csc (x))

n = 1 \sum \infty (\frac{1 + n}{n})^{n}

d er i v a t i v e \frac{- 1}{10 + 4 t ^{2}}

\frac{d}{d x} (e^{x + y} sin (x - y))

d er i v a t i v e (x - 2) (x - 3)^{2}

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