 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 that is e raised to power x into 1 by x minus 1 by x square. So let us start with the solution to this question. We have to find integral i that is equal to integral of e raised to power x into 1 by x minus 1 by x square dx. What we do is we put e raised to power x divided by x equal to t or we can say this is same as e raised to power x into 1 by x equal to t. Now we differentiate both the sides with respect to x. So here we apply the product rule that will be the first function that is e raised to power x into differentiation of second function is minus 1 by x square plus the second function that is 1 by x into differentiation of first function that is e raised to power x is equal to dt by dx. This implies e raised to power x into 1 by x minus 1 by x square dx is equal to dt. This is same as i, this function is same as the function that we have in this integral. So we can write it as i is equal to integral dt and integral dt is equal to t plus a constant c. Now we put back the value of t as e raised to power x divided by x 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.