![\int \limits_1^{\infty}\frac{\ln x}{x^2}\,dx=\lim_{t\to\infty}\int \limits_1^t \frac{\ln x}{x^2}\\\\ \int \frac{\ln x}{x^2}\,dx=(*)\\ u=\ln x,du=\frac{1}{x} \\ dv=\frac{1}{x^2},v=-\frac{1}{x}\\ (*)=\ln x\cdot(-\frac{1}{x})-\int (\frac{1}{x}\cdot(-\frac{1}{x}))\, dx=\\ -\frac{\ln x}{x}+\int\frac{1}{x^2}\, dx=\\ -\frac{\ln x}{x}-\frac{1}{x}+C\\\\ \lim_{t\to\infty}\int \limits_1^t \frac{\ln x}{x^2}=\lim_{t\to \infty}\left[-\frac{\ln x}{x}-\frac{1}{x} \right]_1^t=](/tpl/images/0187/7657/b30cd.png)
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