Answer
The integral converges.
Work Step by Step
We have
$\int_1^{\infty} \dfrac{\ln x dx}{x^2} \ dx=\lim\limits_{R \to \infty} \int_1^{R} \dfrac{\ln x}{x^2} \ dx \\=\lim\limits_{R \to \infty} [-[\dfrac{1}{x}(\ln x+1)]_1^{R} \\=\lim\limits_{R \to \infty}[1- \dfrac{1}{R} (\ln R+1)]\\=1-\lim\limits_{R \to \infty} \dfrac{\ln R}{R}-\lim\limits_{R \to \infty} \dfrac{1}{R}\\=1-0 \\=1$
Hence, the integral converges.
