Question

\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]

Answer 1

\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{\left( x^2 - 2x + 4 \right)}dx\]
\[ = \int\frac{\left( x^3 + 2^3 \right) \left( x - 1 \right)}{\left( x^2 - 2x + 4 \right)}dx \]
\[ = \int\frac{\left( x + 2 \right) \left( x^2 - 2x + 4 \right) \left( x - 1 \right)}{\left( x^2 - 2x + 4 \right)} dx \left[ \therefore a^3 + b^3 = \left( a + b \right) \left( a^2 - ab + b^2 \right) \right]\]
\[ = \int \left( x + 2 \right) \left( x - 1 \right)dx\]
\[ = \int\left( x^2 - x + 2x - 2 \right)dx\]
\[ = \int\left( x^2 + x - 2 \right)dx\]
\[ = \int x^2 dx + \text{∫ x   dx}  - 2 \int1dx\]
\[ = \frac{x^3}{3} + \frac{x^2}{2} - 2x + C\]