Question

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

Answer 1

\[\int\frac{dx}{1 + x - x^2}\]
\[ = \int\frac{- dx}{x^2 - x - 1}\]
\[ = \int\frac{- dx}{x^2 - x + \frac{1}{4} - \frac{1}{4} - 1}\]
\[ = \int\frac{- dx}{\left( x - \frac{1}{2} \right)^2 - \frac{5}{4}}\]
\[ = \int\frac{dx}{\frac{5}{4} - \left( x - \frac{1}{2} \right)^2}\]
\[ = \int\frac{dx}{\left( \frac{\sqrt{5}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}\]
\[\text{ let x }- \frac{1}{2} = t\]
\[ \Rightarrow dx = dt\]
\[Now, \int\frac{dx}{\left( \frac{\sqrt{5}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}\]
\[ = \int\frac{dt}{\left( \frac{\sqrt{5}}{2} \right)^2 - t^2}\]
\[ = \frac{1}{2 \times \frac{\sqrt{5}}{2}} \text{ log } \left| \frac{\frac{\sqrt{5}}{2} + t}{\frac{\sqrt{5}}{2} - t} \right| + C\]

\[= \frac{1}{\sqrt{5}} \text{ log}\left| \frac{\sqrt{5} + 2t}{\sqrt{5} - 2t} \right| + C\]
\[ = \frac{1}{\sqrt{5}} \text{ log } \left| \frac{\sqrt{5} + 2\left( x - \frac{1}{2} \right)}{\sqrt{5} - 2\left( x - \frac{1}{2} \right)} \right| + C\]
\[ = \frac{1}{\sqrt{5}} \text{ log } \left| \frac{\sqrt{5} - 1 + 2x}{\sqrt{5} + 1 - 2x} \right| + C\]