 Hello and welcome to the session. Let's discuss the following question that says integrate the following function Let's now move on to the solution We have to find the integral of the function square root of x squared plus 1 into log x squared plus 1 minus 2 log x upon x to the power 4 Now i is equal to integral square root of x squared plus 1 can be written as We take x squared common here. So we have x squared into 1 plus 1 over x squared into log x squared plus 1 minus log x squared upon x to the power 4 right because m log n is equal to log n to the power m now again i is equal to Taking x squared out of the root so it becomes x into square root of 1 plus 1 over x squared Into log x squared plus 1 minus log x squared over x to the power 4 Into dx Since we are integrating this with respect to x Again i is equal to integral square root of 1 plus 1 over x squared Now this can be written as log x squared plus 1 upon x squared upon x to the power 3 Into dx and Here we have used the rule log m minus log n is equal to log m by n Now again this integral is equal to square root of 1 plus 1 over x squared log 1 plus 1 over x squared into 1 over xq dx right now we put 1 plus 1 over x squared is equal to t now differentiating both sides we have a differentiation of a constant function is 0 and differentiation of 1 over x square is minus 2 over x to the power 3 Into dx is equal to dt so this is equal to minus 2 over x cube dx is equal to dt So dx over x cube is equal to minus dt by 2 so the integral i becomes square root p Into log t Into dt by minus 2 and we take minus sign out of the integral Now we'll integrate this function using by parts and here we take log t as first function and root t as second function So this becomes integral minus 1 by minus 1 by 2 integral root t into log t dt now by using by parts rule Integral of this would be first function as it is Into integral of second function so integral of second function would be t to the power 1 by 2 plus 1 over 1 by 2 plus 1 minus Integral of derivative of first function derivative of log p is 1 over t Into integral of the second function that is t to the power 1 by 2 plus 1 over 1 by 2 plus 1 dt This is further equal to minus 1 by 2 into log t into t to the power 3 by 2 t to the power 3 by 2 over 3 by 2 minus Integral 1 over t into t to the power 3 by 2 upon 3 by 2 dt So the equal to minus 1 by 2 into 2 by 3 t to the power 3 by 2 log t minus Integral 2 by 3 in Into t to the power 3 by 2 minus 1 that is 1 by 2 dt This is further equal to minus 1 by 2 into 2 by 3 t to the power 3 by 2 log t minus 2 by 3 Integral of t to the power 1 by 2 and it would be 3 by 2 upon 3 by 2 plus C This is further equal to minus 1 by 2 Taking 2 by 3 into t to the power 3 by 2 common we have minus 2 by 3 Plus C. This is further equal to minus 1 by 3 t to the power 3 by 2 into log t minus 2 by 3 plus C Now t is 1 plus 1 over x square. So let's substitute to the power 3 by 2 into log 1 plus 1 over x square minus 2 by 3 plus C Hence the required answer is minus 1 by 3 into 1 plus 1 over x square to the power 3 by 2 into log 1 plus 1 over x square minus 2 by 3 plus C This completes the question on the session. Bye for now. Take care. Have a good day