Question

\[\int\frac{1}{1 - x - 4 x^2}\text{  dx }\]

Answer 1

\[\text{ We  have,} \]
\[I = \int\frac{1}{1 - x - 4 x^2}dx\]
\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - - x^2 \frac{x}{4}}dx\]
\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - \left( x^2 + \frac{x}{4} \right)}dx\]
\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - \left\{ x^2 + + \left( \frac{1}{8} \right)^2 - \left( \frac{1}{8} \right)^2 \frac{x}{4} \right\}}dx\]
\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - \left( x + \frac{1}{8} \right)^2 + \frac{1}{64}}dx\]
\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} + - \left( x + \frac{1}{8} \right)^2 \frac{1}{64}}dx\]
\[ = \frac{1}{4}\int\frac{1}{\frac{16 + 1}{64} - \left( x + \frac{1}{8} \right)^2}dx\]

\[ = \frac{1}{4}\int\frac{1}{\left( \frac{\sqrt{17}}{8} \right)^2 - \left( x + \frac{1}{8} \right)^2}dx\]
\[ = \frac{1}{4} \times \frac{1}{2 \times \frac{\sqrt{17}}{8}} \text{ ln }\left| \frac{\frac{\sqrt{17}}{8} + x + \frac{1}{8}}{\frac{\sqrt{17}}{8} - x - \frac{1}{8}} \right| + C .................\left[ \because \int\frac{1}{a^2 - x^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{a + x}{a - x} \right| + C \right]\]
\[ = \frac{1}{\sqrt{17}} \text{ ln }\left| \frac{\frac{\sqrt{17} + 1}{8} + x}{\frac{\sqrt{17} - 1}{8} - x} \right| + C\]