[tex]\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=[/tex]
[tex]\lim_{t\to \infty}\left(-\frac{\ln t}{t}-\frac{1}{t}-\left(-\frac{\ln 1}{1}-\frac{1}{1}\right)\right)=\\
\lim_{t\to \infty}\left(-\frac{(\ln t)'}{t'}\right)-0-(-1)=\\
\lim_{t\to \infty}\left(-\frac{\frac{1}{t}}{1}\right)+1=\\
\lim_{t\to \infty}\left(-\frac{1}{t}\right)+1=\\
0+1=\\
1
[/tex]
