 Hello and welcome to the session. The question says integrate the following function and the 23 is x plus 2 upon root over x square minus 1. Let's start with the solution. We have to integrate the given function. So we have integral x plus 2 upon root over x square minus 1 into dx, which can be written as integral x upon root over x square minus 1 into dx plus integral 2 into dx upon root over x square minus 1. Now, first let us integrate this function. Let us name it as 1 and then we shall integrate the second function. Now, for the first function, let us say x square minus 1 is equal to t. So this implies 2x into dx is equal to dt or x dot dx is equal to dt upon 2. So this first function can be written as dt upon 2 into root over t or we have 1 upon 2 integral dt upon root t and this is equal to 1 upon 2 into 2 times of root t plus c1 where c1 is a constant. So this is further equal to root over x square minus 1 plus c1. So this is the value of the first integral. Now let us solve the second one which is taking 2 outside the integral sign. Again we have dx upon root over x square minus 1 and 1 can be written as 1 square. Now this further equal to 2 into log mod x plus root over x square minus 1 plus c2. Therefore this function which is integral x dot dx upon x square minus 1 root over plus 2 times integral of dx upon root over x square minus 1 can be written as root over x square minus 1 plus c1 plus into log mod x plus root over x square minus 1 plus c2. What is this further equal to root over x square minus 1 plus 2 log mod x plus root over x square minus 1 plus c where c is equal to c1 plus c2. Thus on integrating the given function we get root over x square minus 1 plus 2 log mod x plus root over x square minus 1 plus c. So this completes the session. Bye and take care.