 Hi and welcome to the session. Let's walk out the following question. The question says evaluate Integral x into square root x raised to power 4 minus 1 dx. Let us start with a solution to the question Let I be the integral that is given to us Integral x into square root x raised to power 4 minus 1 into dx. Now here we put x squared to be equal to t. Now differentiating both the sides we get 2x dx is equal to dt. Therefore integral i now becomes 1 by 2 into Integral square root t square minus 1 into dt. Now we know that integral of square root x square minus a square dx is equal to x by 2 into square root x square minus a square minus a by 2 into log mod of x plus square root x square minus a square plus c. So in this we have x to as t and a is 1. So we simply put in the formula here and we get 1 by 2 into t by 2 into square root t square minus 1 minus 1 by 2 into log of mod t plus square root t square minus 1 plus the constant c. Now this can be further written as what we do now is we take 1 by 2 common from both these terms and we get 1 by 4 into t into square root t square minus 1 minus log t plus square root t square minus 1 plus the constant c. Now here we see that with the log we have mod t plus square root t square minus 1. Therefore we can say i is equal to 1 by 4 into x square multiplied by x raised to power 4 minus 1 minus log of mod x square plus square root x raised to power 4 minus 1 plus c. Now what we have done here is we have simply put the value of t as x square. So this is our answer to this question. I hope that you understood the solution. I enjoyed the session. Have a good day.