I can't solve this, no matter how I try

$$\int_{-\infty}^{+\infty} \frac{e^{2x}}{4e^{3x}+9}\,dx$$

Thanks in advance


Hint. You may write $$ \begin{align} \int_{-\infty}^{+\infty} \frac{e^{2x}}{{4e^{3x}+9}}dx &= \int_{-\infty}^{+\infty} \frac{e^{x}}{{4e^{3x}+9}}\times e^{x}dx = \int_0^{+\infty} \frac{u}{{4u^{3}+9}}\,du \end{align} $$ then use a partial fraction decomposition to evaluate the latter integral or use the beta function to get $\displaystyle \frac{\pi }{9 \times2^{1/3} \times3^{1/6}}$ as the desired value.

Some details concerning the use of the beta function. We make the following change of variable $$ v=\frac{9}{{4u^{3}+9}},\quad u=\left(\frac94\right)^{1/3}\left(\frac1v-1\right)^{1/3},\quad du=\left(\frac94\right)^{1/3}\frac13\left(-\frac1{v^2}\right)\left(\frac1v-1\right)^{-2/3}dv $$ giving

$$ \begin{align} \int_{-\infty}^{+\infty} \frac{e^{2x}}{{4e^{3x}+9}}dx &=\int_0^{+\infty} \frac{u}{{4u^{3}+9}}\,du \\\\&= \left(\frac94\right)^{2/3}\frac1{27}\int_0^1 \left(\frac1v-1\right)^{1/3}v\left(\frac1{v^2}\right)\left(\frac1v-1\right)^{-2/3}dv\\\\ &= \left(\frac94\right)^{2/3}\frac1{27}\int_0^1 v^{1/3-1}\left(1-v\right)^{2/3-1}dv\\\\ &=\left(\frac94\right)^{2/3}\frac1{27} \times B(1/3,2/3)\\\\ &=\color{red}{ \frac{\pi }{9 \times2^{1/3} \times3^{1/6}}.} \end{align} $$


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