 Hello and welcome to the session I am Deepika here. Let's discuss the question which says evaluate the integral using substitution integral from 0 to 1 x upon x square plus 1 dx. So let's start the solution. x plus 1 is equal to t in the given integral. Now consider the integral without limit x upon x square plus 1 dx is equal to, therefore x dx is equal to 1 by 2 t is equal to 1 by substitution. We have reduced the given integral to n, the constant of integration. Now we will re-substitute for the new variable that is equal to x square plus 1. So this is equal to 1 by 2 log square plus 1. Hence, from 0 to 1 x upon x square plus 1 dx, now we will find the values of answers, limits of integral. So this is equal to 1 by 2 log of from 0 to 1 the difference of the values, lower limits. So this is equal to 1 by 2 log 2. So we have this is equal to 1 by 2 log 2. So for the above question is 1 by 2 log 2. Hope the solution is clear to you. Bye and take care.