Question

$\int \frac{e^{3x}}{4e^{6x}-9}dx$

Answer 1

$$\begin{gathered} \int \frac{e^{3x}dx}{4e^{6x}-9} \\ \mathrm{let}{e}^{3x}=t \\ \Rightarrow e^{3x}\times 3dx=dt \\ \Rightarrow e^{3x}dx=\frac{dt}{3} \\ \mathrm{Now},\int \frac{e^{3x}dx}{4e^{6x}-9} \\ =\frac{1}{3}\int \frac{dt}{4t^2-9} \\ =\frac{1}{3}\int \frac{dt}{{\left(2t\right)}^2-3^2} \\ =\frac{1}{3}\times \frac{1}{2\times 3}\log \left|\frac{2t-3}{2t+3}\right|\times \frac{1}{2}+C \\ =\frac{1}{36}\log \left|\frac{2t-3}{2t+3}\right|+C \\ =\frac{1}{36}\log \left|\frac{2e^{3x}-3}{2e^{3x}+3}\right|+C \end{gathered}$$