Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=\dfrac{1}{x^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=-\dfrac{1}{x}\end{matrix}\right.\)
\(I=-\dfrac{lnx}{x}|^a_1+\int\limits^a_1\dfrac{dx}{x^2}=-\dfrac{lna}{a}-\dfrac{1}{a}+1\Rightarrow1-\dfrac{1}{a}=\dfrac{1}{2}\Rightarrow a=2\)



