 Hi and welcome to the session. I am Purva and I will help you with the following question. Prove the following integral dx upon x square into x plus 1 where limit is from 1 to 3 is equal to 2 by 3 plus log 2 by 3. Now we begin with this solution. We denote this left hand side by i. So we have i is equal to integral limit is from 1 to 3 dx upon x square into x plus 1. We prove this question by partial fractions method. So we consider 1 upon x square into x plus 1 and this is equal to a upon x plus b upon x square plus c upon x plus 1 where a, b and c are real numbers and we have to find out the values. This implies 1 is equal to a into x into x plus 1 plus b into x plus 1 plus c into x square and this is equal to opening the brackets we get a into x square plus a into x plus b into x plus b plus c x square. Equating the coefficients of x square, x and constant terms we get. Now equating the coefficients of x square on right hand side and left hand side we get a plus c is equal to 0. Equating the coefficients of x on right hand side and left hand side we get a plus b is equal to 0 and equating the constant terms on both the sides we get b is equal to 1. So we have got b is equal to 1. Now putting this value of b in the second equation that is equation 2 we get a plus 1 is equal to 0 which implies a is equal to minus 1. Now putting this value of a in first equation we get minus 1 plus c is equal to 0 which implies c is equal to 1. Now putting the values of a, b and c in this equation star we get putting the values of a, b and c in star we get 1 by x square into x plus 1 is equal to minus 1 by x plus 1 upon x square plus 1 upon x plus 1. Therefore we get integral dx upon x square into x plus 1 where limit is from 1 to 3 is equal to minus integral 1 upon x dx where limit is from 1 to 3 plus integral where limit is from 1 to 3 1 by x square dx plus integral limit is from 1 to 3 dx upon x plus 1. This is equal to now integrating the right hand side we get minus log x because integration 1 by 3 is equal to log x plus into minus 1 by x because integral 1 by x square is equal to minus 1 by x plus now integral 1 by x plus 1 is equal to log x plus 1. In each case we have limit from 1 to 3 this is equal to minus log putting upper limit in place of x we get 3 minus minus log putting lower limit we get 1 plus minus 1 upon putting upper limit we get 3 minus minus 1 upon putting lower limit we get 1 plus log putting upper limit 3 in place of x we get 3 plus 1 is equal to 4 minus log putting lower limit 1 we get 1 plus 1 is 2. So log 2 this is equal to minus log 3 minus into minus becomes plus log 1 minus 1 by 3 minus into minus becomes plus 1 plus log now we can write 4 as 2 square so we get log 2 square minus log 2. Now this is equal to minus log 3 plus log 1 now minus 1 by 3 plus 1 gives plus 2 by 3 and we can write this log 2 square as 2 log 2 minus log 2 and this is equal to minus log 3 now we know that log 1 is 0 so plus 0 plus 2 by 3 plus now 2 log 2 minus log 2 gives log 2. Now this is further equal to we can write this as 2 by 3 plus now log 2 minus log 3 can be written as log 2 by 3 and this is equal to our right hand side. Hence we have proved that integral limit is from 1 to 3 dx upon x square into x plus 1 is equal to 2 by 3 plus log 2 by 3 hope you have understood the solution take care and god bless you.