Chapter 5 - Accumulating Change: Limits of Sums and the Definite Integral - 5.9 Activities - Page 409: 16

$-\dfrac {1}{\ln x}+c$

$v=\ln x\Rightarrow dv=\dfrac {1}{x}dx$ $$\Rightarrow \int \dfrac {1}{x\left( \ln x\right) ^{2}}dx=\int \dfrac {1}{\left( \ln x\right) ^{2}}\times \dfrac {1}{x}dx=\int \dfrac {1}{v^{2}}dv=\int v^{-2}dv=-v^{-1}+c=-({\ln x})^{-1}+c=-\dfrac {1}{\ln x}+c$$