It looks like you are trying to compute the improper integral,
[tex]I = \displaystyle\int_1^\infty \dfrac{\mathrm dx}{x(9+\ln^2(x))}[/tex]
or some flavor of this. If this interpretation is correct, substitute u = ln(x) and du = dx/x. Then
[tex]I = \displaystyle\int_0^\infty \dfrac{\mathrm du}{9+u^2} \\\\ = \frac13\arctan\left(\frac u3\right)\bigg|_{u=0}^{u\to\infty} \\\\ = \frac13\lim_{u\to\infty}\arctan\left(\frac u3\right) \\\\ = \frac13\times\frac\pi2 = \boxed{\frac\pi6}[/tex]
