 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says find the following integral that is integral of x cube minus x square plus x minus 1 divided by x minus 1 into dx. So let us start with the solution to this question. We have to find integral of x cube minus x square plus x minus 1 divided by x minus 1 dx. Now this can be written as integral of x square into x minus 1 plus 1 into x minus 1 divided by x minus 1 into dx. We have written this because we have seen that from taking x square common from these two terms we get x minus 1 plus taking one common from these two terms we get this into x minus 1. So this can be written as integral of x square plus 1 into x minus 1 divided by x minus 1 dx. Now we see that x minus 1 gets cancelled with x minus 1 and we have integral of x square plus 1 dx. Now separating these two terms we get integral of x square dx plus integral of 1 dx. This can be written as x cube by 3 plus x plus c. This happens because we see that integral of x raise to power n dx is equal to x raise to power n plus 1 divided by n plus 1 plus a constant c where n is not equal to minus 1. Now here we see n is 2 so n plus 1 we will have x raise to power 3 divided by 3. 1 is same as x raise to power 0 so n in this case is 0 so we get an x. So our answer to this question is x cube by 3 plus x plus c. So I hope that you understood the question and enjoyed the session. Have a good day.