 Hi and welcome to the session. Let us discuss the following question. Question says integrate the following function. Given function is square root of 1 plus x square upon 9. First upon let us understand that integral of square root of x square plus a square dx is equal to 1 upon 2x multiplied by square root of x square plus a square plus a square upon 2 multiplied by log of x plus square root of x square plus a square plus c. This formula of integration is our key idea to solve the given question. Let us now start with the solution. Now we have to integrate function square root of 1 plus x square upon 9 with respect to x. First upon let us consider the given function. Now this function can be written as square root of 9 plus x square upon 9. Now this is further equal to 1 upon 3 multiplied by square root of 9 plus x square. Now we can write this integral as 1 upon 3 multiplied by integral square root of 9 plus x square dx. Now we know 9 is square of 3 so we can write this integral as 1 upon 3 multiplied by integral square root of x square plus 3 square dx. Now using the formula given in key idea we can write this integral is equal to 1 upon 3 multiplied by 1 upon 2 x multiplied by square root of x square plus 3 square plus 3 square upon 2 multiplied by log of x plus square root of x square plus 3 square plus c where c is the constant of integration. Now this is further equal to x upon 6 multiplied by square root of x square plus 9 plus 3 upon 2 multiplied by log of x plus square root of x square plus 9 plus c. Now this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.