compute the integral

[Infernum]

[tex]\int_{}\frac{1}{\ln x}(1-\frac{1}{\ln x}) dx[/tex]

[ins-en]

This integral is very simple but is nice. My solution is the following:
[tex]\int \frac{1}{lnx}\left(1-\frac{1}{lnx}\right)\,dx[/tex] may be represented in th following way:
[tex]\int \frac{1}{lnx}\,dx - \int \left(\frac{1}{lnx}\right)^{2}\,dx[/tex]
Integration by parts ot the first integral gives that it is equal to:
[tex]\frac{x}{lnx} + C + \int \left(\frac{1}{lnx}\right)^{2}\,dx - \int \left(\frac{1}{lnx}\right)^{2}\,dx[/tex]
So the final answer is: [tex]\frac{x}{lnx} + C[/tex]