 Hello students, let's work out the following problem. It says evaluate the integral 1 upon e to the power x minus 1 So let's now start the solution and let I be the integral 1 upon e to the power x minus 1 dx. Now to solve this integral, we'll put e to the power x as t. So e to the power x into dx is equal to dt So dx is equal to dt upon e to the power x, but e to the power x is t. So it is dt upon t So the integral becomes 1 upon t into t minus 1 dt. Now to solve this integral we'll resolve into partial fractions So we'll put 1 upon p into t minus 1 is equal to a upon t plus b into t minus 1. Now we have to find the values of these constants. So we have a into t minus 1 plus b into t is 1 is equal to a into t minus 1 plus b into t. Now comparing the coefficients of t on both sides we have 0 is equal to a plus b. So this implies a is equal to minus b and Comparing the constant term on both sides we have 1 is equal to minus a. This implies a is equal to minus 1 So this implies v is equal to 1 So 1 upon t into t minus 1 is equal to minus 1 upon t plus 1 upon t minus 1. So the integral i becomes minus 1 upon t plus 1 upon t minus 1 into dt which is again equal to separating the integrals we have 1 upon t dt plus the integral 1 upon t minus 1 dt Now the integral of 1 upon t is log t. So it is since here we have negative sign. So it is minus log t plus log t minus 1 plus c where c is the constant of integration So now t is e to the power x. So it is log e to the power x plus log e to the power x minus 1 plus c Again, this is equal to log e to the power x minus 1 upon e to the power x plus c as we know that log a minus log b is equal to log a by b. So this is the required value of the integral log e to the power x minus 1 upon e to the power x plus c So this completes the question and the session. Bye for now. Take care. Have a good day