Question

\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{  dx }\]

Answer 1

\[\int\left( \frac{x^2 + x - 1}{x^2 + x - 6} \right)dx\]
\[\frac{x^2 + x - 1}{x^2 + x - 6} = 1 + \frac{5}{x^2 + x - 6}\]
\[ \int\left( \frac{x^2 + x - 1}{x^2 + x - 6} \right)dx\]
\[ = ∫ dx + 5\int\frac{dx}{x^2 + x - 6}\]
\[ = ∫  dx + 5\int\frac{dx}{x^2 + x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 - 6}\]
\[ = ∫ dx + 5\int\frac{dx}{\left( x + \frac{1}{2} \right)^2 - \frac{1}{4} - 6}\]

\[ = ∫ dx + 5\int\frac{dx}{\left( x + \frac{1}{2} \right)^2 - \left( \frac{5}{2} \right)^2}\]

\[ = x + 5 \times \frac{1}{2 \times \frac{5}{2}} \text{ log } \left| \frac{x + \frac{1}{2} - \frac{5}{2}}{x + \frac{1}{2} + \frac{5}{2}} \right| + C\]
\[ = x + \text{ log } \left| \frac{x - 2}{x + 3} \right| + C\]