An improper integral of $\ln x/x^2$ by parts

Integration by parts on $[1,a]$. For $a>1$, take $u(x)=\ln x$ and $v'(x)=\frac{1}{x^2}$, so $u'(x)=\frac1x$ and $v(x)=-\frac1x$. Then
$$\int_1^a\frac{\ln x}{x^2}\,dx=\left[-\frac{\ln x}{x}\right]_1^a+\int_1^a\frac{dx}{x^2}.$$

Evaluate.
$$\left[-\frac{\ln x}{x}\right]_1^a=-\frac{\ln a}{a}+0=-\frac{\ln a}{a},\qquad \int_1^a\frac{dx}{x^2}=\left[-\frac1x\right]_1^a=-\frac1a+1.$$

Limit as $a\to+\infty$. Since $\frac{\ln a}{a}\to0$ and $\frac1a\to0$,
$$\int_1^{+\infty}\frac{\ln x}{x^2}\,dx=\lim_{a\to+\infty}\left(-\frac{\ln a}{a}-\frac1a+1\right)=1.$$