 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 minus 3 into e raised to power x divided by x minus 1 the whole cube. So let us start with the solution to this question. Let the integral i be equal to integral of x minus 3 into e raised to power x divided by x minus 1 the whole cube dx. Now this can be written as integral of e raised to power x into x minus 1 minus 2 divided by x minus 1 the whole cube into dx where x minus 1 can be taken as one term and minus 2 as the other term. x minus 1 minus 2 is same as x minus 3. Now when separating this numerator we get integral of e raised to power x into x minus 1 divided by x minus 1 the whole cube minus 2 divided by x minus 1 the whole cube dx. This is equal to integral of e raised to power x. Now we see that on cancelling x minus 1 with one of the x minus 1 in the denominator we get 1 divided by x minus 1 the whole square. This cube becomes square minus 2 divided by x minus 1 the whole cube dx. Now we put e raised to power x into 1 divided by x minus 1 the whole square to be equal to t. So we will have e raised to power x into minus 2 into x minus 1 raised to power minus 3 plus 1 divided by x minus 1 the whole square into e x the whole into dx is equal to dt. Now let us see how do we get this from this. We have differentiated both the sides. Here we have applied the product rule so we will have first function that is e raised to power x into differentiation of first second function. Differentiation of 1 divided by x minus 1 the whole square is minus 2 into x minus 1 raised to power minus 3 plus the second function to differentiation of first function that is e raised to power x will be equal to dt by dx then we have taken dx on the left hand side. So this is how we get this. This implies that e raised to power x into 1 divided by x minus 1 the whole square minus 2 divided by x minus 1 the whole cube dx is equal to dt. Now we have seen that i was also equal to this in this function that is this function. So we can write that i is equal to integral dt integral dt is equal to t plus c. Now we put back the value of t that was e raised to power x divided by x minus 1 the whole square plus c. So our answer to this question is e raised to power x divided by x minus 1 the whole square plus c. So I hope that you understood the question and enjoyed the session. Have a good day.