Question

$\frac{(x^{3} + 8)(x - 1)}{x^{2} - 2x + 4}dx$ equals.

• A. $\left( \frac{x^{3}}{3} \right) + \left( \frac{x^{2}}{2} \right) - 2x + c$
• B. $x^{3} + x^{2} - 2x + c$
• C. $\frac{(x^{3} + x^{2} - x)}{3} + c$
• D. None of these

Answer 1

Answer: (A)

$\left( \frac{x^{3}}{3} \right) + \left( \frac{x^{2}}{2} \right) - 2x + c$

Answer 2

$\int_{}^{}\frac{(x^{3} + 8)(x - 1)}{x^{2} - 2x + 4}dx$

$= \int_{}^{}\frac{(x + 2)(x^{2} - 2x + 4)(x - 1)}{x^{2} - 2x + 4}$

$= \int_{}^{}{(x + 2)(x - 1)}dx$ $= \int_{}^{}{(x^{2} + x - 2})dx$ =$\frac{x^{3}}{3} + \frac{x^{2}}{2} - 2x + c$