 Hello and welcome to the session. The given question says evaluate the following definite integral and the 14th integral x upon 1 plus x square dx and the lower limit of the integral is 2 and the upper limit is 3. So let's start with the solution. First let us find the value of integral x upon 1 plus x square with respect to x. Let us take t is equal to 1 plus x square. So this implies dt is equal to 2x into dx or x into dx is equal to dt upon 2. So this integral can further be written as dt upon 2t taking 1 upon 2 outside the integral sign we have dt upon t in place of 1 plus x square we have substituted t and this is equal to half log t which is half log mod t is 1 plus x square does integral 2 to 3 x upon 1 plus x square into dx can be written as half log mod 1 plus x square from 2 to 3 by the second fundamental theorem of integral calculus which says if we have an integral fx dx from a to b then it's equal to fp minus fa where this f is the entity derivative of the given function f thus now this can be written as half log mod 1 plus x square from 2 to 3 this can further be written as half now the value of this function at the upper limit is given by log 1 plus 3 square minus log now value of this function at the lower limit which is the entity derivative of the given function is log mod 1 plus 2 square now this is further equal to half log 3 square is 9 9 plus 1 is 10 minus log 5 and this is equal to half log 10 by 5 since log a minus log b is equal to log a upon b and I am further simplifying we have 52010 so our answer is half log 2 thus on evaluating the given definite integral our answer is half log 2 so this completes the session hope you have understood it by intake care you