Prove $\int_{\ln(3)}^{+\infty}{\frac x{e^{2x}-9}}dx$ diverges

Question

I need some help proving, that the integral $$\int_{\ln(3)}^{+\infty}{\frac x{e^{2x}-9}}dx$$ diverges. I have tried applying a comparison test $$\frac x{e^{2x}-9}=\frac x{(e^x-3)(e^x+3)}\ge \frac 1{(e^x-3)(e^x+e^x)}=\frac 12\frac 1{(e^x-3)e^x}, for\space x\in(\ln(3), +\infty)$$ which then implies $$\int_{\ln(3)}^{+\infty}{\frac x{e^{2x}-9}}dx\ge\frac 12\int_{\ln(3)}^{+\infty}{\frac 1{(e^x-3)e^x}}dx=\begin{bmatrix}u:=e^x\\du:=e^xdx\end{bmatrix}=\frac 12\int_3^{+\infty}{\frac 1{(u-3)u^2}}du$$ but I can't think of an easy way to prove, that the last integral diverges.

Answer 1

Since$$\lim_{u\to3}\frac{\frac1{(u-3)u^2}}{\frac1{u-3}}=\frac19$$and $\int_3^4\frac1{u-3}\,\mathrm du$ diverges, then so does does $\int_3^4\frac1{(u-3)u^2}\,\mathrm du$ and therefore so does your integral.

Answer 2

HINT:

Note that $$\frac1{u^2(u-3)}=-\frac1{9u}-\frac1{3u^2}+\frac{1}{9(u-3)}$$

Can you proceed now?

Answer 3

$\frac 1{u^2(u-3)}>0$ for $u\in(3;+\infty)$. Therefore, $\int_3^{+\infty}{\frac 1{u^2(u-3)}}du \ge\int_3^4{\frac 1{u^2(u-3)}}du$.

On this interval $\frac 1{u^2(u-3)}\ge\frac 1{16(u-3)}$. This implies, that $$\int_3^4{\frac 1{u^2(u-3)}}du\ge\frac 1{16}\int_3^4{\frac 1{u-3}}du=\frac 1{16}[\ln(u-3)+C]_3^4=-\frac1{16}\lim_{u\to 3^+}{\ln(u-3)}=+\infty$$ From the chain of inequalities we can conclude, that the original integral diverges too.