 Hello and welcome to the session, my name is Asha and I am going to help you with the following question which says, integrate the following functions 19th is 1 upon root over x minus a into x minus b. If we have a polynomial of the type ax square plus bx plus c then it can be written as a into x plus b upon 2 a whole square plus c upon a minus b square upon 4 a square. This is the key idea which we shall be using in this problem to integrate the given function. Now we shall try to write the denominator of this function as the sum of the squares of two numbers. Let us start with the solution. First consider the product of x minus a into x minus b. This is further equal to x square minus a plus b into x plus a into b. Now this can further be written as x plus minus a minus b upon 2 whole square plus a b upon 1 minus a plus b whole square upon 4 into 1 square with the help of our key idea and this is further equal to x minus a plus b upon 2 whole square plus a b minus a square minus b square minus 2 a b and I am taking else here. Here we have 4 a b upon 4. So it can further be written as x minus a plus b upon 2 whole square plus on simplifying this we have minus a square minus b square plus 2 a b upon 4. Taking minus sign outside the bracket this is further equal to x minus a plus b upon 2 whole square minus and this is equal to a square plus b square minus 2 a b which is the formula of a minus b whole square upon 2 whole square and this is further equal to x minus a plus b upon 2 whole square minus a minus b upon 2 whole square. Thus the given function can be written as 1 upon root over x minus a into x minus b is equal to 1 upon root over x minus a plus b upon 2 whole square minus a minus b upon 2 whole square. Now we have to integrate this function that is we have integral 1 upon root over x minus a plus b upon 2 whole square minus a minus b upon 2 whole square into dx. Let us take x minus a plus b upon 2 is equal to t this implies dx is equal to dt by differentiating both sides with respect to x. So this is further equal to integral dt upon root over t square minus a minus b upon 2 whole square. Now if we have a integral of the type 1 upon root over x square minus a square then on integrating with respect to x it is equal to log mod x plus root over x square minus a square plus c. So applying this formula here we have log mod p plus root over p square minus a minus b upon 2 whole square plus c or this can further be written as log x minus a plus b upon 2 plus and the value of this root is equal to x minus a plus b upon 2 whole square minus a minus b upon 2 whole square which is equal to x minus a into x minus b. So here we have root over x minus a into x minus b plus c. Thus on integrating the given function we get log mod x minus a plus b upon 2 plus root over x minus a into x minus b plus c. So this completes the session Bye and take care.