<s>[INST] $ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$ where $a,b \gt 0$ Evaluate
$$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$$ where $a,b \gt 0$ 
I tried using $y=e^x$, but I still can't solve it.
I get $\displaystyle\int_0^\infty \frac y{ay^3+b} \, dy.$
Is there any different method to solve it?
 [/INST] It suffices to compute:
$$\int \frac{y}{y^3 - r^3}dy$$
(putting $r = - \sqrt[3]{\frac{b}{a} }$). Let's assume that $a, b > 0$.
After decomposing into partial fractions, i.e.:
$$\frac{y}{y^3 - r^3} = \frac1{3r} \left( \frac1{y - r} - \frac{y - r}{y^2 + ry + r^2} \right)$$
we find:
$$\int\frac{y}{y^3 -r^3} dy = \frac1{3r}\left[ \log\left( \frac{y-r}{\sqrt{y^2 + ry + r^2}} \right) + \sqrt 3 \arctan \left( \frac{2y + r}{r \sqrt 3} \right) \right]$$
Hence, 
$$\int_0^{\infty} \frac{y}{y^3-  r^3} dy = \frac{1}{r\sqrt 3} \left( \arctan (-\infty) - \arctan\left( \frac1{\sqrt 3} \right) \right)=  -\frac{2\pi}{3r\sqrt 3}$$
and so,
$$\int_{-\infty}^{\infty} \frac{e^{3x}}{ae^{2x} + b} dx  = \frac{2\pi}{3\sqrt 3 \sqrt[3]{a^2 b}}$$
</s>