 Hello and welcome to the session. Let us discuss the following question, question says, find the following integrals. Here we have to find the integral of x square multiplied by 1 plus 1 upon x square dx. First of all let us understand that integral of fx plus dx dx is equal to integral of fx dx plus integral of dx dx and integral of x raised to the power n dx is equal to x raised to the power n plus 1 upon n plus 1 plus c where c is any constant also integral of dx is equal to x plus c where c is any constant. Now we will use these expressions as our key idea to solve the given question. Let us now start with the solution. Now we have to find integral of x square multiplied by 1 minus 1 upon x square dx. Now multiplying x square by this bracket we get integral of x square minus 1 dx. Now using the expression 1 given in the key idea we can write this integral as integral of x square dx minus integral of dx. From expression 2 of the key idea we know integral of x square is equal to x raised to the power 2 plus 1 upon 2 plus 1 and integral of dx is x and c is any constant. Now this expression is further equal to x cube upon 3 minus x plus c. So we get integral of x square multiplied by 1 minus 1 upon x square dx is equal to x cube upon 3 minus x plus c. This is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.