Integral of 1 Over x lnx

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

The integration 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, on taking differentiation, we have dz = \frac{1}{x}dx, so the given integral (i) takes of 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 again 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}