 Hello students, let's solve the following problem of integration. We have to integrate the function x plus 1 into x plus log x whole squared upon x. Let us now proceed on with the solution and let I be the integral x plus 1 into x plus log x whole squared upon x. Here we see that the derivative of x plus log x is 1 plus 1 upon x which can be written as x plus 1 upon x that is this So put equal to x plus log x. So dt by dx is equal to 1 plus 1 upon x taking lcm this becomes x plus 1 upon x. So this implies dt is equal to x plus 1 upon x dx. So x plus 1 upon x into dx is dt and x plus log x is t. So substituting all these values in the integral the integral i becomes t squared dt and the integral of t squared dt is t to the power 2 plus 1 upon 2 plus 1 plus c which is equal to p cube upon 3 plus c. This is by the formula of the integral of x to the power n which is equal to x to the power n plus 1 upon n plus 1 plus c and here n is 2. Now t is x plus log x. So substitute it. So this becomes x plus log x cube upon 3 plus c as the integral of the given function is 1 by 3 into x plus log x whole cube plus c. And this completes the question and the session. Bye for now. Take care. Have a good day.