 Hi and welcome to our session. Let us discuss the following question. The question says, integrate the following functions e to the power 3 log x into x to the power 4 plus 1 to the power minus 1. Let's now begin with this illusion. Let i is equal to integral of e to the power 3 log x into x to the power 4 plus 1 to the power minus 1 with respect to x. This is equal to integral of e to the power log x cube into x to the power 4 plus 1 to the power minus 1 with respect to x. We know that e to the power log x is equal to x so this is equal to integral x cube into x to the power 4 plus 1 to the power minus 1 which can be written as 1 by x to the power 4 plus 1 dx 4 plus 1 as t. Now this implies dt is equal to dx is 1 by 4 into dt is equal to x cube dx. So by substituting t in place of x to the power 4 plus 1 we get to dt into right mod here because x to the power 4 is always positive and