 Hello students, let's solve the following problem of integration. We have to integrate the function x cube minus 1 whole to the power 1 by 3 into x to the power 5. Let us now proceed on with the solution and let i be the integral x cube minus 1 whole to the power 1 by 3 x to the power 5 dx. Now we see that the derivative of x cube minus 1 is 3x square which contains x square. So to get the term of x square we break x to the power 5 as x to the power 3 into x to the power 2. So it becomes x cube minus 1 whole to the power 1 by 3 into x to the power 3 into x to the power 2 dx. Now what equal to x cube minus 1? So dt by dx is equal to 3x square and this implies dt is equal to 3x square dx. And we need to have x square dx so we divide both sides by 3. So dt by 3 is equal to x square dx. So x square dx is equal to dt by 3 and t is equal to x cube minus 1. So let us substitute all these values. So also if t is equal to x cube minus 1 so x cube is equal to t plus 1. So substituting all these values in the integral i integral becomes t to the power 1 by 3 because x cube minus 1 is t into x cube which is t plus 1 which is further written as t to the power 1 by 3 into t plus t to the power 1 by 3. Multiplying t to the power 1 by 3 with t and 1 and it is dt by 3 because x square dx is equal to dt by 3 which is again equal to t to the power 4 by 3 plus t to the power 1 by 3 dt by 3. And it is further written as 1 by 3 integral of t to the power 4 by 3 dt plus 1 by 3 integral t to the power 1 by 3 dt. Now the integral of t to the power 4 by 3 is given by t to the power 4 by 3 plus 1 upon 4 by 3 plus 1 plus 1 by 3 into the integral of t to the power 1 by 3 which is 1 by 3 plus 1 upon 1 by 3 plus 1 plus c which is the constant of integration of both the integrals. This is again equal to 1 by 3 into t to the power 7 by 3 upon 7 by 3 plus 1 by 3 into t to the power 4 by 3 upon 4 by 3 plus c. Now 3 gets cancelled with 3. We have 1 by 7 t to the power 7 by 3 plus 1 by 4 into t to the power 4 by 3 plus c. Now t is x cube minus 1 so let us substitute as 1 by 7 x cube minus 1 whole to the power 7 by 3 plus 1 by 4 to t to the power 4 by 3 t is x cube minus 1 4 by 3 plus c. And here we have used the formula for the integration of x to the power m dx which is equal to x to the power n plus 1 upon n plus 1 plus c. Hence the integral of the given function is 1 by 7 into x cube minus 1 whole to the power 7 by 3 plus 1 by 4 into x cube minus 1 whole to the power 4 by 3 plus c. This completes the question and the session. Hope you enjoy this session goodbye and take care.