 Hello and welcome to the session, the question says, integrate the following function 28 as x plus 3 upon x square minus 2x minus 5. So, let us start with the solution and we have to integrate the given function with respect to x. So, let us take x plus 3 is equal to 8 times the derivative of x square minus 2x minus 5 with respect to x plus p. This is further equal to x plus 3 is equal to 8 derivative of this function is 2x minus 2 plus p. Now equating the coefficients of the variable x and the constants we have a is equal to 1 upon 2 and 3 is equal to minus 2a plus b which implies on simplifying that b is equal to 4. That is the given integral which is x plus 3 upon x square minus 2x minus 5 into dx can be written as integral. Now x plus 3 is equal to a into 2x minus 2 plus p and a is half into 2x minus 2 plus 4 upon x square minus 2x minus 5 into dx. So, in place of x plus 3 we have put a into 2x minus 2 plus b and in place of a and b we have put their numerical values. Now this is further equal to now separating both the integrals and taking constants outside the integral we have half integral 2x minus 2 into dx upon x square minus 2x minus 5 plus 4 times integral dx upon x square minus 2x minus 5. Now let us solve both these integrals one by one. The first one is integral 2x minus 2 upon x square minus 2x minus 5 into dx. Let us put t is equal to x square minus 2x minus 5. So, this implies dt is equal to 2x minus 2 into dx. So, this integral can further be written as dt upon t and this is equal to log mod p plus the constant c1 and this is further equal to log mod t is x square minus 2x minus 5 plus c1. So, the value of the first integral is half log mod x square minus 2x minus 5 plus c1 and we have plus 4 and now let us solve this integral. So, we have integral dx upon x square minus 2x minus 5. Now, if we have a polynomial of the type A x square plus B x plus c then it can be written as A into x plus B upon 2a whole square plus c upon A minus B square upon 4a square. So, writing this polynomial which is x square minus 2x minus 5 this way it is equal to x minus 1 whole square plus then we have c upon A, c is minus 5, A is 1. So, we have minus B square is minus 2 whole square this is equal to 4 upon 4 into 1 square. So, this is further equal to x minus 1 whole square minus 5 plus 1 is 6. So, we have root 6 whole square. So, this function can further be written as integral dx upon x minus 1 whole square minus root 6 whole square. Now, let us take x minus 1 is equal to t. So, this implies dx will be equal to dt. So, this integral can further be written as dt upon t square minus root 6 whole square. And the formula to integrate the function of the type dx upon root x square minus A square is equal to 1 upon 2a log mod x minus a upon x plus a plus a constant. So, this can be written as 1 upon 2 root 6 log mod t minus root 6 upon t plus root 6 plus a constant c 2. So, this is further equal to 1 upon 2 root 6 log value of t is x minus 1 minus root 6 upon x minus 1 plus root 6 plus c 2. So, here the value of this integral is equal to 1 upon 2 root 6 log mod x minus 1 minus root 6 upon x minus 1 plus root 6 plus a constant c 2, which is further equal to half log mod x square minus 2x minus 5 plus 2 into 2 is 4. So, we have 2 upon root 6 log mod x minus 1 minus root 6 upon x minus 1 plus root 6 plus a constant c, where c is the sum of c 1 and c 2. Hence, the answer is half log mod x square minus 2x minus 5 plus 2 upon root 6 log mod x minus 1 minus root 6 upon x minus 1 plus root 6 plus a constant c. So, this completes the presentation. Bye and take care.