 Hi and welcome to the session. Let us discuss the following question. Question says, integrate the following functions. Given function is 3x square upon x raised to the power 6 plus 1. First of all, let us understand that integral dx upon x square plus a square is equal to 1 upon a tan inverse x upon a plus c, where c is the constant of integration. Now we will use this formula as our key idea to solve the given question. Let us now start with the solution. You have to find integral of 3x square upon x raised to the power 6 plus 1 with respect to x. Now this integral can be further written as integral of 3x square dx upon square of x cube plus 1. Now we know 3x square is derivative of x cube. So we will substitute x cube is equal to t. So we can write put x cube is equal to t. Now differentiating both the sides with respect to x, we get 3x square is equal to dt upon dx. Now this further implies 3x square dx is equal to dt. Let us name these expressions as 1 and 2. Now we can write integral 3x square upon x raised to the power 6 plus 1 dx is equal to integral 3x square upon square of x cube plus 1 dx. Now using expression 1 and 2 this integral is equal to integral dt upon t square plus 1. Now using formula of the integral given in the key idea we get this integral is equal to tan inverse t plus c. Clearly we can see variable x has been replaced by t and a is equal to 1. So we get 1 upon a tan inverse t. 1 upon a is 1 only as a is equal to 1. So we get tan inverse t plus c where c is the constant of integration. Now t is equal to x cube so substituting x cube for t we get tan inverse x cube plus c. So we get integral of 3x square upon x raised to the power 6 plus 1 dx is equal to tan inverse x cube plus c. This is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.