Use definite integration by parts

Take $u=\ln x$, $dv/dx=1$, so $du/dx=1/x$ and $v=x$. Thus $\int_1^2\ln x\,dx=[x\ln x]_1^2-\int_1^2x(1/x)\,dx$.

This is $(2\ln2-0)-[x]_1^2=(2\ln2-0)-(2-1)=2\ln2-1$.