Integral of 1 Over x lnx

In this tutorial we shall find the integration of 1 over x lnx function. To evaluate this integral we shall use the method of substitution of integration.

The integration is of the form
\[\begin{gathered} I = \int {\frac{1}{{x\ln x}}dx} \\ I = \int {\frac{1}{{\ln x}}\frac{1}{x}dx} \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered} \]

To solve this integration, putting $$z = \ln x$$ and differentiating, we have $$dz = \frac{1}{x}dx$$, so the given integral (i) takes the form
\[I = \frac{1}{z}dz\]

Using the formula of integration $$\int {\frac{1}{x}dx = \ln x + c} $$
\[I = \ln z + c\]

By putting the value $$z = \ln x$$ in the evaluated integral, we have
\[\begin{gathered} I = \ln \left( {\ln x} \right) + c \\ \Rightarrow \int {\frac{1}{{x\ln x}}dx} = \ln \left( {\ln x} \right) + c \\ \end{gathered} \]