I've been working on \begin{align} \int_0^\infty\frac{1}{x(x+1)(x+2)(x+4)-x(x+1)e^{x-0.5x^2}} dx. \end{align}
I struggle to simplify the denominator. Does anyone have a suggestion how I could continue with the integral?
I tried something like partial fractions. Let $\mathcal{P}(x)=(x+2)(x+4)$ and $\mathcal{E}(x)=e^{x-0.5x^2}$. Then, $$\frac{1}{x(x+1)(\mathcal{P}(x)-\mathcal{E}(x))}=\frac{1}{x(x+1)}-\frac{\mathcal{P}(x)-\mathcal{E}(x)-1}{x(x+1)(\mathcal{P}(x)-\mathcal{E}(x))}$$ and the first term is easy now but I'm again not sure how to proceed.
I'm grateful for @Henry Lee's suggestion to do a substitution which changes the exponential to a single quadratic term, $e^{-0.5x^2}$, similar to how it appears in the Gaussian bell curve. But this didn't help me to greatly simplify further calculations.
this is not a full answer but a few suggestions: $$I=\int\frac{dx}{x(x+1)(x+2)(x+4)-x(x+1)e^{x-0.5x^2}}=\int\frac{1}{x(x+1)}\frac{dx}{(x+2)(x+4)-e^{x-0.5x^2}}$$ now we can see that: $$x-\frac{x^2}{2}=-\frac12\left(x^2-2x\right)=-\frac12\left[(x-1)^2-1\right]$$ so potentially with a substitution $u=x-1$ we end up with: $$I=\int\frac{1}{(u+1)(u+2)}\frac{du}{(u+4)(u+6)-e^{1/2}e^{-u^2/2}}$$