 Hi, and welcome to the session. Let's work out the following question. The question says evaluate integral dx divided by square root x multiplied by 1 plus square root x. So let's start with a solution to this question. So let i be equal to integral dx divided by square root x into 1 plus square root x. We put 1 plus square root x to be equal to t. We differentiate both the sides and we get 1 upon 2 square root x dx is equal to dt. On cross multiplication we get dx is equal to or we can say that 1 upon square root x into dx is equal to 2 dt. Therefore integral i becomes 2 into integral dt divided by t that is further equal to 2 log mod t plus a constant c. We put back the value of t as 1 plus square root x and we get 2 into log of mod 1 plus square root x plus the constant c. So this is our answer to this question. I hope that you understood the solution and enjoyed the session. Have a good day.