Integral of ln(x)/x^2
If we do as I said and let $\displaystyle \ln(x)\longmapsto z$ then
$\displaystyle \begin{aligned}\int\frac{\ln(x)}{dx}&\overbrace{\l ongmapsto}^{\ln(x)=z}\int\frac{z}{e^{2z}}\cdot e^z dz\\
&=\int z\cdot e^{-z}dz\\
&=-z\cdot e^{-z}-e^{-z}\quad{\color{red}\star}\\
&=\frac{-\ln(x)}{x}-\frac{1}{x}+C\end{aligned}$
$\displaystyle \color{red}\star$ was gotten using integration by parts with $\displaystyle u=z$