 Hi and welcome to the session. I am Purva and I will help you with the following question. Integrate the function 1 upon x raise to the power 1 by 2 plus x raise to the power 1 by 3. Now we begin with the solution. We denote this function by i. So we have i is equal to integral 1 upon x raise to the power 1 by 2 plus x raise to the power 1 by 3 dx. This is equal to integral 1 upon norm multiplying and dividing the denominator by x raise to the power 1 by 3 we get x raise to the power 1 by 3 into x raise to the power 1 by 6 plus 1 right dx. Now we put x raise to the power 1 by 6 as t. So we have put x raise to the power 1 by 6 equal to t or we can write this as x is equal to t raise to the power 6. Now differentiating we get dx is equal to 6 into t raise to the power 5 dt. Also x raise to the power 1 by 3 is equal to t raise to the power 6 by 3. Now cancelling out common factors 3 numerator and denominator in power we get this is equal to t square. So putting all these values in i we get i is equal to integral 6 into t raise to the power 5 upon t square into t plus 1 dt. Now cancelling out t square in denominator and numerator we get this is equal to integral 6 into t cube upon t plus 1 dt. This is equal to integral now we take out 6 and we divide t cube by t plus 1. So we get the numerator as t plus 1 into t square minus t plus 1 minus 1 upon t plus 1 dt. This is equal to 6 integral t plus 1 into t square minus t plus 1 upon t plus 1 dt minus 6 integral dt upon t plus 1. Cancelling out t plus 1 in denominator and numerator we get this is equal to 6 integral t square minus t plus 1 dt minus 6 integral dt by t plus 1. We can write this as this is equal to 6 integral t square dt minus 6 integral dt plus 6 integral dt minus 6 integral dt upon t plus 1. This is equal to 6 into integrating t square we get t cube by 3 minus 6 into integrating t we get t square by 2 plus 6 integrating dt we get t minus 6 into integrating 1 by t plus 1 we get log t plus 1 plus c where c is the constant. Cancelling out 3 here in denominator and numerator we get 2 and here cancelling out 2 in denominator and numerator we get 3. So this is equal to 2 into t cube minus 3 into t square plus 6 t minus 6 into log 1 plus t plus c. This is equal to 2 into now we know that t is equal to x raise to the power 1 by 6. So x raise to the power 1 by 6 whole raise to the power 3 minus 3 into x raise to the power 1 by 6 whole raise to the power 2 plus 6 into x raise to the power 1 by 6 minus 6 into log 1 plus x raise to the power 1 by 6 plus c. Now cancelling out 3 and 6 here we get 2 in denominator and cancelling out 2 and 6 here we get 3 in denominator. So we get this is equal to 2 into x raise to the power 1 by 2 minus 3 into x raise to the power 1 by 3 plus 6 into x raise to the power 1 by 6 minus 6 into log 1 plus x raise to the power 1 by 6 plus c. So we get the answer as 2 into root x minus 3 into x raise to the power 1 by 3 plus 6 into x raise to the power 1 by 6 minus 6 into log 1 plus x raise to the power 1 by 6 plus c. So this is the answer of the question. Hope you have understood the solution. Take care and God bless you. .