 Hello and welcome to the session. The question says integrate the following function and the 17th one is 1 upon root over x minus 1 into x minus 2. So first let us learn how do we change the polynomial of the form x square plus bx plus c as the sum of the squares of two polynomials. This is equal to a into x plus b upon 2a whole square plus c upon a minus b square upon 4a square. This is on page number 614 of your book which says to find the integral of the form whose denominator is a polynomial. So there we try to write the denominator which is in the form of a x square plus bx plus c as the sum of the squares of two polynomials. So this is our p idea with the help of which we shall integrate the given function. Let us now start with the solution. So here we have given the function 1 upon root over x minus 1 into x minus 2. Now x minus 1 into x minus 2 is equal to x square minus 3x plus 2 which can with the help of this written as x plus minus 3 upon 2 whole square plus 2 upon 1 minus 9 upon 4 into 1 square which is further equal to x minus 3 upon 2 whole square plus 2 minus 9 upon 4 which is further equal to x minus 3 upon 2 whole square plus on simplifying this bracket we have minus 1 upon 4 which is equal to on further simplifying x minus 3 upon 2 whole square minus 1 upon 2 whole square. Thus the given function will be written as 1 upon root over x minus 3 upon 2 whole square minus 1 upon 2 whole square. Now we have to integrate this function. So we have integral 1 upon root over x minus 3 upon 2 whole square minus 1 upon 2 whole square into dx. Let us put x minus 3 upon 2 is equal to t. So the search dx is equal to dt and thus we can further write it as dt upon root over t square minus half square. Now the formula to integrate the function of the type root over 1 upon root over x square minus a square is equal to log x plus root over x square minus a square plus c. So this can be written as log more t plus root over t square minus half square plus c where c is the constant and t is equal to x minus 3 upon 2. So this can further be written as log mod 2 x minus 3 upon 2 plus t square minus half square is equal to x minus 3 upon 2 whole square minus half whole square and this is equal to x minus 1 into x minus 2. Thus here we can write it as integral x minus 1 into x minus 2 plus c and this is further equal to log mod x minus 3 upon 2 plus root over on solving this we have x square minus 3 x plus 2 plus c. Thus on integrating the given function we get log mod x minus 3 upon 2 plus root over x square minus 3 x plus 2 plus c. So this completes the session hope you have understood it take care and have a good day.