 Hi and welcome to our session. I'm Kanika and I'm going to help you to solve the following question. The question says integrate the following functions. Given function is e to the power x by 1 plus e to the power x into 2 plus e to the power x. Now begin with this illusion. Let i is equal to integral of e to the power x by 1 plus e to the power x into 2 plus e to the power x with respect to x. The integrand is a rational function in e to the power x. Therefore, we will first put e to the power x as t. Now e to the power x equals to t implies tt is equal to e to the power x into tx. Now substituting t in place of e to the power x, we get i as integral of 1 plus t into 2 plus t with respect to t. Now this is a proper rational function so we will now resolve it into partial fractions. So let 1 by 1 plus t into 2 plus t is equal to a by 1 plus t plus b by 2 plus t where a and b are real numbers which we have to determine. Now 1 by 1 plus t into 2 plus t equals to a by 1 plus t plus b by 2 plus t implies 1 is equal to a into 2 plus t plus b into 1 plus t. Now put t equals to minus 1. Now by substituting t as minus 1 we get 1 equals to a into 2 minus 1. This implies a is equal to 1. Now we will find the value of t. So now we are going to put t as minus 2. By substituting minus 2 in place of t we get 1 equals to e into 1 minus 2. This implies b is equal to minus 1. So value of a is 1 and value of b is minus 1. By 1 plus t into 2 plus t is equal to 1 by 1 plus t minus 1 by 2 plus t. So integral of 1 by 1 plus t into 2 plus t with respect to t is equal to integral of 1 by 1 plus t with respect to t minus integral of 1 by 2 plus t with respect to t. We know that integral of 1 by ax plus b with respect to x is equal to log mod ax plus b divided by derivative of ax plus b that is a plus c. So using this integral of 1 by 1 plus t with respect to t is equal to log mod 1 plus t minus integral of 1 by 2 plus t with respect to t is equal to log mod 2 plus t plus c. We know that log a minus log b is equal to log a by b. So using this log mod 1 plus t minus log mod 2 plus t can be written as log mod 1 plus t by 2 plus t plus c. We have assumed t as e to the power x. So by replacing t by e to the power x we get log 1 plus e to the power x by 2 plus e to the power x plus c. We don't need to write mod here because e to the power x is always positive and so 1 plus e to the power x and 2 plus e to the power x is also positive. Hence our required answer is log 1 plus e to the power x by 2 plus e to the power x plus c. So this completes the session. Bye and take care.