Răspuns:
Explicație pas cu pas:
[tex]I=\int\limits^1_e {\frac{lnx}{x^{2}}} \, dx =\int\limits^1_e {\frac{1}{x^{2}}lnx} \, dx\\f=lnx-->f^{,} =\frac{1}{x} \\g^{,} =\frac{1}{x^{2} } -->g=-\frac{1}{x}\\I=-\frac{1}{x}lnx |^{e}_{1} -\int\limits^1_e {\frac{1}{x}(-\frac{1}{x})} \, dx=-\frac{1}{e}+\int\limits^1_e {\frac{1}{x^{2}}} \, dx=-\frac{1}{e}-\frac{1}{x} |^{e}_{1}=-\frac{1}{e}-\frac{1}{e}+1=-\frac{2}{e}+1=\frac{e-2}{e}[/tex]
