 Hi, and how are you all today? The question says evaluate integral 1 upon log x minus 1 upon log x the whole square tx. Here we are given the integral as 1 upon log x minus 1 upon log x the whole square dx. Right? Now here, let us put log x equal to t or x to the power t or dx equal to e raised to the power t d t. Right? And on doing so, we have now our integral like integral 1 upon t minus 1 upon t square e raised to the power tt. Now, let us open the brackets i is equal to integral e raised to the power t upon t d t minus integral e raised to the power t upon t square dt. Now here on using my parts, we have the first function into integral of second minus integral derivative of the first function into integral of the second function. Let us make this let us not integrate this. Further implies i is equal to bracket e raised to the power t upon t minus minus 1 upon t square into e raised to the power t minus integral e raised to the power t upon t square dt plus c1 plus c2. Further implies i is equal to e raised to the power t upon t plus e raised to the power t upon t square minus all these were in integral sign, sorry minus integral e raised to the power t upon t square dt plus c1 plus c2. These two will get cancelled out and we are left with i is equal to e raised to the power t upon t plus c. c1 plus c2 has been combined and written as c. Now on substituting like the value of i and t, we have e raised to the power t was x upon t was log x plus c. So this is the required answer to the given question. So hope you understood it well and have a nice day.