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Question
\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]
Sum
Solution
\[\int\frac{e^{3x} dx}{4 e^{6x} - 9}\]
\[\text{let }e^{3x} = t\]
\[ \Rightarrow e^{3x} \times 3dx = dt\]
\[ \Rightarrow e^{3x} dx = \frac{dt}{3}\]
\[Now, \int\frac{e^{3x} dx}{4 e^{6x} - 9}\]
\[ = \frac{1}{3}\int\frac{dt}{4 t^2 - 9}\]
\[ = \frac{1}{3}\int\frac{dt}{\left( 2t \right)^2 - 3^2}\]
\[ = \frac{1}{3} \times \frac{1}{2 \times 3} \text{ log }\left| \frac{2t - 3}{2t + 3} \right| \times \frac{1}{2} + C\]
\[ = \frac{1}{36} \text{ log }\left| \frac{2t - 3}{2t + 3} \right| + C\]
\[ = \frac{1}{36} \text{log }\left| \frac{2 e^{3x} - 3}{2 e^{3x} + 3} \right| + C\]
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