The given function is \( \displaystyle \dfrac{(x^3 + 1)(x \; – 2)}{x^2 \; – x \; – 2} \) and we need to find the integration of this function with respect to \( x \).
Solution:
\( \begin{align*} &\int \dfrac{(x^3 + 1)(x \,- 2)}{x^2 \,- x \,- 2} \, dx \\ &= \int \dfrac{(x^3 + 1^3)(x \,- 2)}{x^2 \,- x \,- 2} \, dx \\ &= \int \dfrac{(x + 1)(x^2 \,- x + 1)(x \,- 2)}{x^2 \,- 2x + x \,- 2} \, dx \;[\because a^3+b^3 = (a+b)(a^2+b^2-ab)] \\ &= \int \dfrac{(x + 1)(x^2 \,- x + 1)(x \,- 2)}{(x \,- 2)(x + 1)} \, dx \\ &= \int \left( x^2 \,- x + 1 \right) \, dx \\ &= \int x^2 \, dx \,- \int x \, dx + \int 1 \, dx \\ &= \dfrac{x^3}{3} \,- \dfrac{x^2}{2} + x + C, \quad \text{C is a constant}\\ \end{align*} \)Thus, the integration of the given function is
\( \boxed{\int \dfrac{(x^3 + 1)(x \,- 2)}{x^2 \,- x \,- 2} \, dx = \dfrac{x^3}{3} \,- \dfrac{x^2}{2} + x + C} \\ \)Please let me know in the comments if you find any errors in this solution.