evaluate the integral ∫(x³ - 2)/(x² - x - 2) dx.

Explanation:

Step1: Factor the denominator

First, factor $x^{2}-x - 2=(x - 2)(x+1)$.

Step2: Perform polynomial long - division

Divide $x^{3}-2$ by $x^{2}-x - 2$. We have $x^{3}-2=(x + 1)(x^{2}-x - 2)+3x$. So, $\frac{x^{3}-2}{x^{2}-x - 2}=x + 1+\frac{3x}{x^{2}-x - 2}$.

Step3: Decompose the fraction into partial fractions

Let $\frac{3x}{x^{2}-x - 2}=\frac{3x}{(x - 2)(x + 1)}=\frac{A}{x - 2}+\frac{B}{x + 1}$. Then $3x=A(x + 1)+B(x - 2)$. Let $x = 2$, we get $A = 2$. Let $x=-1$, we get $B = 1$. So, $\frac{3x}{x^{2}-x - 2}=\frac{2}{x - 2}+\frac{1}{x + 1}$.

Step4: Integrate term - by - term

$\int\frac{x^{3}-2}{x^{2}-x - 2}dx=\int(x + 1)dx+\int\frac{2}{x - 2}dx+\int\frac{1}{x + 1}dx$. $\int(x + 1)dx=\frac{x^{2}}{2}+x+C_1$, $\int\frac{2}{x - 2}dx=2\ln|x - 2|+C_2$, $\int\frac{1}{x + 1}dx=\ln|x + 1|+C_3$.

Answer:

$\frac{x^{2}}{2}+x + 2\ln|x - 2|+\ln|x + 1|+C$ (where $C = C_1+C_2+C_3$)