 Hello and welcome to the session. The question says, integrate the following function. 10th is x square upon root over x square plus a square. So first let us learn a formula to integrate the function of the diet 1 upon root over x square plus a square with respect to x. So this is equal to log mod x plus root over x square plus a square plus a. So with the help of this formula we shall integrate the given function. So this is our D idea. Now let's start with the solution. The given function is x square upon root over x raise to the power 6 plus a raise to the power 6. It can also be written as x square upon root over x cube raise to the power 2 plus a cube raise to the power 2. Now we have to integrate this function that is integral x square dot dx upon root over x cube whole square plus a cube whole square. This is further equal to, first let us substitute sorry put x cube is equal to t. So this implies 3x square into dx is equal to dt which further implies that x square into dx is equal to dt upon 3. So this integral can further be written as dt upon 3 into root over t square plus a cube whole square. Taking the constant outside the integral we have 1 upon 3 integral dt upon root over t square plus a cube whole square. Now with the help of this key idea this can further be written as 1 upon 3 log mod x and raise to the power x we have t plus root over t square plus a cube whole square that is a raise to the power 6 plus c where c is a constant. Now let us put the value of t which is x cube so we have x cube plus root over x cube whole square plus a raise to the power 6 plus c which is further equal to 1 upon 3 log mod x cube plus root over x raise to the power 6 plus a raise to the power 6 plus c. Thus on integrating the given function we get 1 upon 3 log mod x cube plus root over x raise to the power 6 plus a raise to the power 6 plus c. So this completes the session hope you have understood it take care and have a good day.