 Hello and welcome to the session. In this session we discussed the following question which says evaluate integral x square upon 1 plus x cube dx. Now let's move on to the solution. Consider integral x square upon 1 plus x cube dx. Now here let us assume that 1 plus x cube be equal to t. So on differentiating both the sides we get 3x square dx. That is, this gives us x square dx is equal to 1 upon 3 dt. Therefore from the above integral we get integral 1 upon 3 dt over t. That is on substituting 1 plus x cube equal to t in this integral and taking x square dx equal to 1 upon 3 dt we get this. So this is equal to 1 upon 3 integral dt over t and this is further equal to 1 upon 3 into log modulus t plus c. Now here we take t equal to 1 plus x cube. So this is equal to 1 upon 3 log modulus 1 plus x cube plus c. Thus we have got integral x square dx over 1 plus x cube is equal to 1 upon 3 log modulus 1 plus x cube plus c where this c is the constant of integration. Thus our final answer is integral x square over 1 plus x cube dx is equal to 1 upon 3 log modulus 1 plus x cube plus c where this c is constant of integration. So this completes this session. Hope you have understood the solution for this question.