|
|
Giải
Đặt $u = \ln x\,\,\, \Rightarrow \,\,\,du = \frac{{dx}}{x}$
$dv = \frac{{dx}}{{{x^3}}}\,\,\,\,\, \Rightarrow \,\,\,\,v = \frac{{ - 1}}{{2{x^2}}}$
$I = - \frac{{\ln x}}{{2{x^2}}} + \frac{1}{2}\int\limits_{}^{} {\frac{{dx}}{{{x^3}}} = } - \frac{{\ln x}}{{2{x^2}}} - \frac{1}{{4{x^2}}} + C,\,\,\,\,\,\,\,\,\,\,\,\,C \in R $
Vậy $I = - \frac{1}{{4{x^2}}}\left( {2\ln x + 1} \right) + C$
|