 Hello and welcome to this session. Let us discuss the following question. It says integrate the following function. The given function is 1 upon x minus under the root x. Let us now move on to the solution. We have to integrate 1 upon x minus under the root x dx. Now, this integral can be written as 1 upon x can be written as under the root x into under the root x minus under the root x. And this can again be written as 1 upon under the root x into under the root x minus 1 dx. Now, we see that the derivative of root x minus 1 is 1 by 2 into root x. So, we put y is equal to under the root x minus 1. So, dy by dx is equal to 1 by 2 into x to the power minus 1 by 2. That is dy by dx is equal to 1 by 2 into under the root x. And this implies dy is equal to 2 into under the root x dx. And this implies 2 into dy is 1 upon root x dx. So, 1 upon root x into dx is equal to 2 dy and root x minus 1 is y. Substituting all these values in the integral, the integral becomes 1 upon y into 2 dy. This is equal to 2 into the integral of 1 upon y dy. Now, the integral of 1 upon y dy is log y. So, this is equal to 2 into log y plus c where c is the constant of the integral. Now, substitute the value of y here. It becomes 2 into log root x minus 1 plus c. Hence, the integral of the given function is 2 into log root x minus 1 plus c. And this completes the question. Bye for now. Take care. Have a good day.