\(F'\left(x\right)=\left(-x^{-2}\right)'=\frac{2}{x^3}=\frac{\frac{2}{x^2}}{x}\Rightarrow f\left(x\right)=\frac{2}{x^2}\)
\(I=\int f'\left(x\right)lnx.dx\)
Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=f'\left(x\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{dx}{x}\\v=f\left(x\right)\end{matrix}\right.\)
\(\Rightarrow I=f\left(x\right).lnx-\int\frac{f\left(x\right)}{x}dx=\frac{2lnx}{x^2}+\frac{1}{x^2}+C\)