Answer
Convergent
Work Step by Step
Use u-substitution method.
Plug $u=\ln x $ and $ du=\dfrac{1}{x} dx$
Now, $\int_1^\infty \dfrac{\dfrac{1}{x}}{ \sqrt {1+(\ln^2 x)}} dx=\int_{0}^\infty \dfrac{1}{(1+u^2)} du$
This implies that $\lim\limits_{k \to \infty} [\tan^{-1} (u)]_{0}^k=\lim\limits_{k \to \infty}[(\tan^{-1} (k) -\tan^{-1} (0)]$
so, $\dfrac{\pi}{2}-0=\dfrac{\pi}{2}$
Thus, the series is convergent.