 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says integrate the following function x square log x. So let us start with the solution to this question. We have to find integral x square log x dx. Now according to the I-late rule we see that log x becomes the first function because this is a logarithmic function and x square becomes the second function because this is an algebraic function and logarithmic function is given preference over algebraic function. We call it I. So I will be equal to integration of second function that is x square dx into log x minus integration of dy dx of log x into integral x square dx. Now this is equal to log of mod x into integral x square dx is x cube by 3 minus integral dy dx of log x is 1 by x integral of x square dx is x cube by 3 and this into dx. Now this can be written as log of mod x into x cube by 3 minus. Now this x gets cancelled with one of the x and the numerator we have x square now. So we have minus 1 by 3 integral x square dx. Now this is equal to log of mod x into x cube by 3 minus 1 by 3 into integral of x square dx is x cube by 3 plus a constant c. This is equal to x cube by 3 into log of mod x minus x cube by 9 plus the constant c. So our answer to this question is x cube by 3 into log of mod x minus x cube by 9 plus c. I hope that you understood the question and enjoyed the session. Have a good day.