$ \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?

Dividing both numerator and denominator by $e^{3x}$, we get $$\frac{e^{2x}}{ae^{3x}+b} dx=\frac{e^{-x}}{a+be^{-3x}} dx$$ 
If you use $y=e^{-x}$, then you get $$-\frac{dy}{a+by^3}=-\frac{1}{b}\cdot\frac{dy}{y^3+r^3}$$ where $r=\sqrt[3]{\frac{b}{a}}$
Now if you substitute $y=r\tan^\frac{2}{3} \theta$ and $dy=\frac{2}{3}r\tan^{-\frac{1}{3}} \theta \sec^2 \theta \, d\theta$, your integrand will become
$$-\frac{1}{b}\cdot\frac{\frac{2}{3}r\tan^{-\frac{1}{3}} \theta \sec^2 \theta \, d\theta}{r^3\sec^2 \theta}=-\frac{2}{3br^2}\cdot\tan^{-\frac{1}{3}} \theta \, d\theta$$
If you change the limits properly, then I think you will be able to complete the integration. Hope this helps.
