 Hello and welcome to the session I am Deepika here. Let's discuss the question which says integrate the following rational function 1 over e raise to power x minus 1. Now we know that if the form of rational function is px plus q upon x minus a into x minus b where a is not equal to b, then form of the partial fraction is a upon x minus a plus b upon x minus b where b are real numbers determined suitably. So this is a key idea behind that question. We will take the help of this key idea to solve the above question. So let's start the solution. Again this 1 over e raise to power x minus 1. Now put e raise to power a x is equal to t. Therefore e raise to power x dx is equal to dt or dx is equal to dt upon e raise to power x. Now e raise to power x is t. So we have dx is equal to dt upon t. Now integral of 1 over x minus 1 dx is equal to integral of plus 1 dt according to our key idea 1 over t minus 1 is equal to a over t plus b over t minus 1. So we have 1 is equal to a raise to t minus 1 plus by creating the coefficients of the constant term we get 0 is equal to a plus b. Let us get this equation as number 1 and 1 is equal to minus a. Let us get this equation as number 2. From 2 we have a is equal to minus 1, equation 1 implies minus 1 plus b is equal to 0, v is equal to 1. So we have a is equal to minus 1 and b is equal to 1 plus 1 over t into t minus 1 is equal to a over t that is minus 1 over t plus b over t minus 1 that is 1 over t minus 1. Therefore, integral of 1 over e raise to power x minus 1 dx is equal to plus 1 into integral of 1 over e raise to power x dx plus 1 over e raise to power x minus 1 dx is equal to log minus 1 plus c over that log n minus log n is equal to log of m is equal to minus 1 upon e raise to power x sub of question is log of minus 1 upon e raise to power x plus c. I hope the solution is clear to you. Bye and