 Hello and welcome to the session, the question says integrate the following function, 9th is x minus 1 upon root over x square minus 1. So first let us learn the formula to integrate the function of the type 1 upon x square minus a square root over with respect to x. So this is equal to log mod x plus root over x square minus a square plus c. So with the help of this formula we shall integrate the given function. So this is our D idea. Now let us start with the solution. Here the given function is x minus 1 upon root over x square minus 1 and we have to integrate this function. So we have integral x minus 1 into dx upon root over x square minus 1. Now separating the numerator we have integral x upon root over x square minus 1 dot dx minus dx upon root over x square minus 1 can be written as 1 square. Now first let us solve this integral. But x square minus 1 is equal to t. So this implies 2x dot dx is equal to dt which further implies that x dot dx is equal to dt upon 2. So this integral can be written as dt upon 2 into root over t. This is further equal to half taking the constant outside the integral we have 1 upon root over t. This is further equal to half integral t raise to the power minus half dot dt also sorry. Now we have to integrate this function with respect to t. So this is equal to half t raise to the power minus half plus 1 upon minus half plus 1 plus a constant c. So this is equal to half root over t upon 1 by 2 plus c which is equal to root over t plus c and t we have assumed as x square minus 1. So integral of dt upon 2 root over t which is equal to root t plus c can be written as integral x dot dx upon root over x square minus 1 is equal to root over x square minus 1 plus c. Let us name this expression as 1. So 1 is equal to integral x dot dx upon root over x square minus 1 minus integral 1 dot dx upon root over x square minus 1 plus c. So the value of the first integral is root over x square minus 1 plus c and solving this with the help of the key idea we have log mod of x plus root over x square minus a square plus root over x square minus a square minus a square minus 1 plus c dash where c and c dash are the two constants. So this can further be written as root over x square minus 1 minus log mod x plus root over x square minus 1 plus d where d is a constant which is equal to c plus c dash. Thus on integrating the given function we get root over x square minus 1 minus log mod x plus root over x square minus 1 plus a constant d or we can also write the constant as any alphabet. So in the book it is given to you as c and as c and d both are constants therefore the value of the integral do not change. So this completes the session. Hope you have understood it. Take care and bye for now.