Question

\[\int\frac{1}{3 x^2 + 13x - 10} \text{ dx }\]

Answer 1

\[\int\frac{1}{3 x^2 + 13x - 10}dx\]
\[ = \frac{1}{3}\int\frac{1}{x^2 + \frac{13}{3}x - \frac{10}{3}}dx\]
\[ = \frac{1}{3}\int\frac{1}{x^2 + \frac{13 x}{3} + \left( \frac{13}{6} \right)^2 - \left( \frac{13}{6} \right)^2 - \frac{10}{3}}dx\]
\[ = \frac{1}{3}\int\frac{1}{\left( x + \frac{13}{6} \right)^2 - \frac{169}{36} - \frac{10}{3}}dx\]
\[ = \frac{1}{3}\int\frac{1}{\left( x + \frac{13}{6} \right)^2 - \frac{169 - 120}{36}}dx\]
\[ = \frac{1}{3}\int\frac{1}{\left( x + \frac{13}{6} \right)^2 - \left( \frac{17}{6} \right)^2}dx\]
\[ = \frac{1}{3} \times \frac{1}{2 \times \frac{17}{6}} \text{ ln } \left| \frac{x + \frac{13}{6} - \frac{17}{6}}{x + \frac{13}{6} + \frac{17}{6}} \right|  .............\left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]
\[ = \frac{1}{17} \text{ ln}\left| \frac{x - \frac{2}{3}}{x + 5} \right| + C\]
\[ = \frac{1}{17} \text{ ln }\left| \frac{3x - 2}{3x + 15} \right| + C\]