 Hi and welcome to the session. Let us discuss the following question. Question says, integrate the functions 1 plus log x whole square upon x. Let us now start with the solution. Now we have to integrate the function 1 plus log x whole square upon x. Now we know derivative of 1 plus log x is equal to 1 upon x. So we will use the substitution 1 plus log x is equal to t. Now differentiating both the sides with respect to x we get 1 upon x is equal to dt upon dx. Or we can say dx upon x is equal to dt. Now we can write integral 1 plus log x whole square upon x dx is equal to integral t square dt. We know 1 plus log x is equal to t. So here we can write t square and dx upon x is equal to dt. So we can write dt for dx upon x. Now integrating this function we get t cube upon 3 plus c, where c is the constant of integration. We know formula for integral of x raise to the power n dx is equal to x raise to the power n plus 1 upon n plus 1 plus c, where c is the constant of integration. Here value of n is 2. So integrating this function we get t raise to the power 2 plus 1 upon 2 plus 1. 2 plus 1 is equal to 3. So we get t cube upon 3 plus c. Now we know t is equal to 1 plus log x. So substituting 1 plus log x for t in this expression we get cube of 1 plus log x upon 3 plus c. So our required answer is integral of square of 1 plus log x upon x dx is equal to 1 upon 3 multiplied by 1 plus log x whole cube plus c. This completes the session. Hope you understood the solution. Take care and have a nice day.