$ \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?

$$
ay^3 + b = a\left( y^3 + \frac b a \right)
$$
$$
y^3 + \frac b a = y^3 +c^3 = (y+c)(y^2 - yc + c^2) \quad\text{where } c = \sqrt[3]\frac b a.
$$
So use partial fractions to get
$$
\frac \bullet {y+c} + \frac {((\bullet\, y) + \bullet)} {y^2 - yc + c^2}.
$$
To integrate the second term, you have
$$
\frac {ey+f}{y^2 - yc + c^2},
$$
and letting $u=y^2-yc+c^2$ so $du = (2y-c)\,dy$. Multiplying both sides of the last equality by $e/2$, we get
$$
\frac e 2 \, du = \left( ey+\frac{ce}{2} \right) \, dy.
$$
So
$$
(ey+f)\,dy = \left( \underbrace{\left( ey + \frac{ce}2 \right)\,dy} + \left( f - \frac{ce}2 \right) \, dy  \right).
$$
The substitution handles the part over the $\underbrace{\text{underbrace}}.$  Then you have
$$
\int \frac{dy}{y^2-yc+c^2}. 
$$
Complete the square
$$
y^2 - yc + c^2 = \left( y^2 - yc + \frac{c^2} 4 \right) + \frac{3c^2} 4 = \left( y - \frac c 2 \right)^2 + \frac{3c^2} 4.
$$
So when you integrate, you get an arctangent.
