Question

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

Answer 1

\[\int\frac{dx}{2 x^2 - x - 1}\]
\[ = \frac{1}{2}\int\frac{dx}{x^2 - \frac{x}{2} - \frac{1}{2}}\]
\[ = \frac{1}{2}\int\frac{dx}{x^2 - \frac{x}{2} + \left( \frac{1}{4} \right)^2 - \left( \frac{1}{4} \right)^2 - \frac{1}{2}}\]
\[ = \frac{1}{2}\int\frac{dx}{\left( x - \frac{1}{4} \right)^2 - \frac{1}{16} - \frac{1}{2}}\]
\[ = \frac{1}{2}\int\frac{dx}{\left( x - \frac{1}{4} \right)^2 - \left( \frac{1 + 8}{16} \right)}\]
\[ = \frac{1}{2}\int\frac{dx}{\left( x - \frac{1}{4} \right)^2 - \left( \frac{3}{4} \right)^2}\]
\[\text{ let x } - \frac{1}{4} = t\]
\[ \Rightarrow dx = dt\]

\[Now, \frac{1}{2}\int\frac{dx}{\left( x - \frac{1}{2} \right)^2 - \left( \frac{3}{4} \right)^2}\]
\[ = \frac{1}{2}\int\frac{dt}{t^2 - \left( \frac{3}{4} \right)^2}\]
\[ = \frac{1}{2}\int\frac{dt}{t^2 - \left( \frac{3}{4} \right)^2}\]
\[ = \frac{1}{2 \times \frac{3}{4}} \times \frac{1}{2} \text{ log }\left| \frac{t - \frac{3}{4}}{t + \frac{3}{4}} \right| + C\]
\[ = \frac{2}{3} \times \frac{1}{2} \text{ log }\left| \frac{x - \frac{1}{4} - \frac{3}{4}}{x - \frac{1}{4} + \frac{3}{4}} \right| + C\]
\[ = \frac{2}{3} \times \frac{1}{2} \text{ log } \left| \frac{x - 1}{x + \frac{1}{2}} \right| + C\]
\[ = \frac{1}{3} \text{ log }\left| \frac{2\left( x - 1 \right)}{2x + 1} \right| + C\]
\[ = \frac{1}{3} \left[ \text{ log }\left| \frac{\left( x - 1 \right)}{2x + 1} \right| + \text{ log }\left| 2 \right| \right] + C\]
\[ = \frac{1}{3} \text{ log }\left| \frac{x - 1}{2x + 1} \right| + \frac{1}{3} \text{ log }\left| 2 \right| + C\]
\[ = \frac{1}{3} \text{ log }\left| \frac{x - 1}{2x + 1} \right| + C' \left[ \because C' = \frac{1}{3} \text{ log } \left| 2 \right| + C \right]\]