 Hello friends. Let's discuss the following question. It says integrate the following function. The given function is x squared upon 2 plus 3 x cube whole cube. Let us now proceed on with this solution and let I be the integral x squared upon 2 plus 3 x cube whole cube dx. Now, how we see that the derivative of 2 plus 3 x cube is 9 x square which contains x square. So we put 2 plus 3 x cube equal to t. So dt by dx is equal to 9 x square. So this implies dt is equal to 9 x square dx and this implies x square dx is equal to dt by 9. Now x square into dx is dt by 9 and 2 plus 3 x cube is t. So substituting all these values in the integral, the integral i becomes 1 upon t cube dt upon 9, which is again equal to 1 by 9 integral t to the power minus 3 dt and the integral of this is 1 by 9 into t to the power minus 3 plus 1 upon minus 3 plus 1 plus c, where c is the constant of integration. So this is equal to 1 by 9 into t to the power minus 2 upon minus 2 plus c. Now t is 2 plus 3 x cube. So substitute it. We have minus 1 upon 18 2 plus 3 x cube to the power minus 2 plus c, which is again equal to minus 1 upon 18 into 2 plus 3 x cube whole square plus c. Hence the integral of the given function is minus 1 upon 18 into 2 plus 3 x cube whole square plus c. And this completes the question and the session. Bye for now. Take care and have a good day.