 Hello and welcome to the session. In this session we discuss the following question which says evaluate integral of log x upon x dx. Let's proceed with the solution now. We suppose that i be equal to the integral log x upon x dx and we are supposed to find the value of this i for this we suppose that log x be equal to t. Now differentiating both the sides we get 1 upon x dx is equal to dt. Since we know that dy dx of log x is equal to 1 upon x dx therefore we get i is equal to integral of dt. Since we know that log x is t and 1 upon x dx is dt so i is equal to integral of t dt. So now integrating further we get i is equal to t square upon 2 plus c where the c is the constant of integration. Now as we had assumed log x to be equal to t so in place of p here we put log x so we get i is equal to log x be whole square upon 2 plus c since we have log x is equal to t. So therefore we have the given integral log x upon x dx is equal to log x be whole square upon 2 plus c c being the constant of integration. So this is our final answer. This completes the session. Let me have understood the solution of this question.