 Hi and welcome to the session. Let us discuss the following question. Question says evaluate the following integrals. Given integral is 1 upon x minus 1 upon 2x square multiplied by e raise to the power 2x dx. Lower limit of the given integral is 1 and upper limit of the given integral is 2. Let us now start with the solution. Now we have to evaluate the integral from 1 to 2 1 upon x minus 1 upon 2x square multiplied by e raise to the power 2x dx. Now this integral can be written as integral from 1 to 2 e raise to the power 2x multiplied by 1 upon x dx minus integral from 1 to 2 e raise to the power 2x multiplied by 1 upon 2x square dx. Now first of all let us find out this integral. Clearly we can see this integral consists of product of two functions. Let us name this function as first function and this function as second function. Now integral of product of two functions is equal to first function that is 1 upon x multiplied by integral of second function that is e raise to the power 2x minus integral of e raise to the power 2x upon 2 multiplied by minus 1 upon x square dx. This is integral of derivative of the first function multiplied by integral of the second function. We will write this integral as it is that is 1 upon 2 integral from 1 to 2 e raise to the power 2x upon x square dx. Now we know integral of e raise to the power 2x is equal to e raise to the power 2x upon 2. So we can write this integral as 1 upon x multiplied by e raise to the power 2x upon 2 and limits are from 1 to 2. Here minus and minus sign will get multiplied and we can write this integral as 1 upon 2 multiplied by integral of e raise to the power 2x upon x square dx minus 1 upon 2 multiplied by integral from 1 to 2 e raise to the power 2x upon x square dx. Clearly we can see these are two same integrals having different signs. So these integrals will cancel each other and now substituting the limits we get e raise to the power 2 multiplied by 2 upon 2 multiplied by 2 minus e raise to the power 2 multiplied by 1 upon 2 multiplied by 1. This is the upper limit. Here we have substituted upper limit for x in this function and here we have substituted lower limit that is 1 in this function. Now this is further equal to e raise to the power 4 upon 4 minus e square upon 2. Now subtracting these two terms by taking their LCM we get e raise to the power 4 minus 2 e square upon 4. This is further equal to e square multiplied by e square minus 2 upon 4. So the given integral from 1 to 2 1 upon x minus 1 upon 2x square e raise to the power 2x dx is equal to e square multiplied by e square minus 2 upon 4. This is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.