 Hello friends. Let's work out the following problem. It says integrate the following function. The given function is 1 upon x plus x into log x. Before finding the integral of this function, we first need to know the formula for the integral of 1 upon y dy. It is log y plus c. So this will be the key idea. Let us now move on to the solution. We have to find the integral of 1 upon x plus x log x dx, which is again equal to the integral of 1 upon. We take x common. So we have x into 1 plus log x dx. Now here we see that the derivative of 1 plus log x is 1 upon x. That is this. So we put y is equal to 1 plus log x. Now we differentiate this with respect to y, with respect to x. So dy by dx is equal to 1 upon x. And this implies dy is equal to 1 upon x dx. Now 1 upon x into dx is dy. And 1 plus log x is y. So we substitute all this in the integral. So the integral becomes 1 upon y dy. Now the integral of 1 upon y dy is log y plus c, where c is the constant of integration. Now we substitute the value of y in this. And y is 1 plus log x. So this becomes log 1 plus x, 1 plus log x plus c. The integral of the given function is log 1 plus log x plus c. And this completes the question. Bye for now. Take care. Have a good day.