Question: For exam preparations we got some integrals to practice on but we can figure out whether or not this integral exists or not:

$$\int\_e^x \frac{dt}{\sqrt{(t\ln(t^3))^2-9t^2}}$$

Since it is an exam question it should be easily shown via Direction Comparison Test, Integral Test or any other method, but we can't figure it out. Help would be greatly appreciated.

Answer: 
Assuming that $x>e$, look at
$$t^2 \log ^2\left(t^3\right)-9 t^2$$ and use Taylor series around $t=e$. This gives
$$t^2 \log ^2\left(t^3\right)-9 t^2=18 e (t-e)+36 (t-e)^2+O\left((t-e)^3\right)$$
$$\sqrt{t^2 \log ^2\left(t^3\right)-9 t^2 }=3 \sqrt{2 e} \sqrt{t-e}+3 \sqrt{\frac{2}{e}} (t-e)^{3/2}+O\left((t-e)^{5/2}\right)$$
$$\frac 1{\sqrt{t^2 \log ^2\left(t^3\right)-9 t^2 }}=\frac{1}{3 \sqrt{2 e} \sqrt{t-e}}-\frac{\sqrt{t-e}}{3 \left(\sqrt{2}
 e^{3/2}\right)}+O\left((t-e)^{3/2}\right)$$

So, compare with $???$
Assuming $x>e$ we may utilize the properties of logarithms to write
$$
I=\int\_e^x \frac{\mathrm dt}{\sqrt{(t\ln(t^3))^2-9t^2}}=\int\_e^x \frac{\mathrm dt}{3t\sqrt{\ln^2 t-1}}=\int\_e^x \frac{\mathrm dt}{3t\sqrt{\ln t+1}\sqrt{\ln t-1}}.
$$
The integrand is positive for all $x>e$ so we can then bound $I$ from above with
$$
I\leq \int\_e^x \frac{\mathrm dt}{3e\sqrt{2}\sqrt{\ln t-1}}=\frac{1}{3e\sqrt{2}}\int\_0^{x-e} \frac{\mathrm du}{\sqrt{\ln(u+e)-1}}.
$$
The only problematic point is $u=0$. What can we say about the integrability of $1/\sqrt{\ln(u+e)-1}$ in the neighborhood of $u=0$?