 Hello and welcome to the session. My name is Mansi and I'm going to help you with the following question. The question says integrate the following function that is x log 2x. So let us start with the solution to this question. We have to find integral x log 2x. Now since we see that this is a product of two functions, so we use bypass. Using bypass we can say that this will be the log 2x will be the first function and x will be the second function because as per the islet rule, logarithmic function is given preference over the trigonometric function or over the algebraic function. Since x is an algebraic function and l that is logarithmic, log x is logarithmic function, so log 2x becomes the first function. Here we will have dx. So let us find out this now. We call this i. i will be equal to, as we've seen in earlier questions, this will be equal to first function that is log 2x into integral x dx minus integral d by dx of log 2x into integral of x dx. This will be equal to log 2x into integral of x dx is x square by 2 because we know integral x raised to power n is equal to x raised to power n plus 1 by n plus 1 and in this case is 1 so we get this minus integral of d by dx of log 2x will be 1 by 2x into 2 because d by dx of log 2x will be 1 by 2x into derivative of 2x is 2 that will get multiplied into integral of x dx is again x square by 2 dx. This is equal to x square by 2 log 2x minus, now we see 2 gets cancelled with 2x gets cancelled with 1 of dx so we have 1 by 2 integral x dx this is equal to x square by 2 log of mod 2x minus 1 by 2 into integral of x dx is x square by 2 plus a constant c. We see that we put a mod sign here because we see that log is not defined for negative values so even here we put a mod so we have log of mod 2x so we have this now this can be written as x square by 2 log of mod 2x minus x square by 4 plus c. So, this is our answer to the question I hope that you understood the question and enjoyed the session have a good day.