 Hello and welcome to the session. Let us discuss the following question. It says integrate the following function. The given function is 1 upon x into log x to the power m and here x is greater than 0. Let us now proceed on with the solution. We have to find the integral of 1 upon x into log x to the power n. Now here we see that the derivative of log x is 1 upon x. So we put y is equal to log x. So dy is equal to 1 upon x into dx. So 1 upon x into dx is dy and y is log x. So substituting all these values in the integral, the integral becomes 1 upon y to the power m dy and again this integral becomes y to the power minus m dy. So this integral is equal to y to the power minus m plus 1 upon minus m plus 1 plus c. As the integral of y to the power n dy is equal to y to the power n plus 1 upon n plus 1 plus c. So n is minus m. So the integral is y to the power minus m plus 1 upon minus m plus 1 plus c. Now we substitute the value of y here. y is log x. So this becomes log x to the power 1 minus m upon 1 minus m plus c. Hence the integral of the given function is log x to the power 1 minus m upon 1 minus m plus c. And this completes the question. Bye for now. Take care. Have a good day.