Question

\[\int\frac{1}{4 x^2 + 12x + 5} dx\]

Answer 1

\[\int\frac{dx}{4 x^2 + 12x + 5}\]
\[ = \frac{1}{4}\int\frac{dx}{x^2 + 3x + \frac{5}{4}}\]
\[ = \frac{1}{4}\int\frac{dx}{x^2 + 3x + \left( \frac{3}{2} \right)^2 - \left( \frac{3}{2} \right)^2 + \frac{5}{4}}\]
\[ = \frac{1}{4}\int\frac{dx}{\left( x + \frac{3}{2} \right)^2 - \frac{9}{4} + \frac{5}{4}}\]
\[ = \frac{1}{4}\int\frac{dx}{\left( x + \frac{3}{2} \right)^2 - 1^2}\]
\[\text{ let x} + \frac{3}{2} = t\]
\[ \Rightarrow dx = dt\]
\[Now, \frac{1}{4}\int\frac{dx}{\left( x + \frac{3}{2} \right)^2 - 1^2}\]
\[ = \frac{1}{4}\int\frac{dx}{t^2 - 1^2}\]
\[ = \frac{1}{4} \times \frac{1}{2 \times 1} \text{ log }\left| \frac{t - 1}{t + 1} \right| + C\]
\[ = \frac{1}{8} \text{ log }\left| \frac{x + \frac{3}{2} - 1}{x + \frac{3}{2} + 1} \right| + C\]
\[ = \frac{1}{8} \text{ log }\left| \frac{x + \frac{1}{2}}{x + \frac{5}{2}} \right| + C\]
\[ = \frac{1}{8} \text{ log }\left| \frac{2x + 1}{2x + 5} \right| + C\]