\int^{\infty}_{\ln2}\frac{dt}{t^2}=\int^{\infty}_{\ln2}t^{-2} dt=\frac{t^{-1}}{-1}|^{\infty}_{\ln2}=-\frac{1}{t}|^{\infty}_{\ln2}=-(\frac{1}{\infty}-\frac{1}{\ln2})=\frac{1}{\ln2}