 Hello and welcome to the session, the question says, integrate the following function a thus x square upon 1 minus x raise to the plus x. So, first let us learn the formula to integrate the function of the form 1 upon a square minus x square with respect to x is equal to 1 upon 2 a into log mod a plus x upon a minus x plus c. So, with the help of this formula we are going to integrate the above question. So, this is our key idea. Now, let us start with the solution. The given function is x square upon 1 minus x raise to the plus x. Now, we have to integrate this function. So, we have integral x square upon 1 minus x raise to the plus x into dx which can further be written as integral x square dot dx upon 1 square minus x cube whole square. Now, let us put x cube is equal to t. So, this implies that 3x square into dx is equal to dt by integrating both sides with respect to x. This further implies that x square into dx is equal to dt upon 3. So, this is further equal to integral dt upon 3 into 1 square minus x cube whole square taking the constant outside the integral saying we have 1 upon 3 integral dt upon 1 square minus t square since x cube we have assumed as t. Now, let us apply the formula with the help of sake idea 1 upon 3 into integral dt upon 1 square minus t square into written as 1 upon 2 a square a is 1. So, 2 into 1 into log and we have a plus x upon a minus x and a is 1 and x is t. So, the numerator we have 1 plus t in the denominator we have 1 minus t the constant c. So, it can further be written as 1 upon 6 log cube. So, we have 1 plus x cube upon 1 minus x cube plus c. Thus, on integrating the given function we get 1 upon 6 log mod 1 plus x cube upon 1 minus x cube plus c. So, this completes the session by