 Hello and welcome to the given session. It says evaluate the following definite integral 13th integral dx upon x square minus 1, lower limit of integration is 2 and upper limit is 3. So let's start with the solution. First let us find the value of integral dx upon x square minus 1 by the method of partial fraction. Now let 1 upon x square minus 1 is equal to a upon x plus 1 plus b upon x minus 1. So let's implies 1 is equal to a plus b into x plus b minus a, when equating the coefficients of constants in x on both side, this implies a plus b is equal to 0 and b minus a is equal to 1. So we have b is equal to a plus 1. So from here we have a plus b is equal to 0 and b is a plus 1. So substituting the value of b here we have a plus a plus 1 is equal to 0 which implies 2 a is equal to minus 1 or a is equal to minus 1 by 2. So a is minus 1 by 2 and a plus b is equal to 0. So this implies b is equal to minus of a that is minus of minus half which is equal to half. Therefore we have a is equal to minus half and b is equal to half. So 1 upon x square minus 1 can be written as in place of a which is substituted minus half and in place of b which is substituted half. So we have minus half upon x plus 1 plus half upon x minus 1. Thus integral dx upon x square minus 1 can be written as minus half integral dx upon x plus 1 plus half integral x upon x minus 1. Now let us solve these two integrals one by one. First let us solve integral dx upon x plus 1. So let us take t is equal to x plus 1. So this implies dt is equal to dx. So here we have integral dt upon t and this is equal to log mod t and t is x plus 1. So we have log mod x plus 1. So this is the value of first integral. So here we have minus half log mod x plus 1. Now let us find the value of second integral which is integral dx upon x minus 1. Now here also let t is equal to x minus 1. So similarly again we have dt is equal to dx. So this can be written as integral dt upon t which is equal to log mod t. Now this is log mod x minus 1. So here we have plus half log mod x minus 1. Now this can further be written as taking half common. We have log mod x minus 1 minus log mod x plus 1 is equal to half log mod x minus 1 upon x plus 1. Since log a minus log b is equal to log a upon b. Now by second fundamental theorem of integral calculus the given definite integral can be written as the given definite integral is 2 to 3 dx upon x square minus 1. So it can be written as half log mod x minus 1 upon x plus 1 lower limit is 2 and upper limit is 3. Since by the second fundamental theorem of integral calculus we have integral fx dx lower limit of integration a and upper limit of integration t is equal to the value of the entire derivative of the given function at the point b minus the value of the entire derivative of the same function at the point a. So here this is the entire derivative. Now let us put the upper limit then put a minus sign and then we will find the lower limit of the entire derivative at the point 2. Taking half common inside we have log mod 3 minus 1 upon 3 plus 1 minus log mod 2 minus 1 upon 2 plus 1. This is further equal to half log 2 upon 4 minus log 1 upon 3 which can further be written as half log 2 by 4 upon 1 by 3 since log a minus log b is log a upon b. So we further have log 2 by 4 into 3 upon 1. On cancelling we have half log 3 upon 2. Thus on evaluating the given integral our answer is half log 3 upon 2. So this completes the session. Hope you have enjoyed it. Take care and bye for now.