I tried 2 tests, but both failed.
Test-1:
checking the integrand, $\log{x}\le x$ in the domain, $\frac{\log{x}}{x^2} \le \frac{x}{x^2} = \frac{1}{x^2}$ and $\int_1^\infty\frac{1}{x}\,dx$ is divergent by p-series test, but we cant say anything about the smaller integral.
Test-2:
Splitting the given integral as $\int_1^\infty$=$\int_1^3$+$\int_3^\infty$. The first term is a proper integral and hence finite, the second can be solved by considering fact that $\log{x}>1, \forall x \ge 3$, and hence, $\frac{\log{x}}{x^2} \ge \frac{1}{x^2}$ ($x\ge 3$) and we have $\int_3^\infty\frac{1}{x^2}\,dx$ convergent by p-series test, but we cant say anything about the larger integral.
So please help on how to approach this problem.
Best Answer
Since you already know that $\log x \leqslant x$, use $\log x = 2 \log x^{1/2}\leqslant 2 x^{1/2}$ to get the estimate
$$\frac{\log x}{x^2}\leqslant \frac{2x^{1/2}}{x^2}= \frac{2}{x^{3/2}}$$
The presence of a constant is irrelevant in applying the comparison test.