 Hello men, welcome to the session. It says integrate the following function 25 is x plus 2 upon root over 4x minus x square. So let's start with the solution and we have to integrate this function. So we have integral x plus 2 upon root over 4x minus x square into dx. Now its numerator x plus 2 can be written as a into derivative of the function 4x minus x square with respect to x plus b. This implies that x plus 2 is equal to a into 4 minus 2x plus b or we have minus 2a is equal to 1 and 4a plus b is equal to 2. So this implies a is equal to minus half and from here b is equal to 2 plus 4 upon 2 which is equal to 4 plus the integral x plus 2 upon root over 4x minus x square can be written as integral minus half into 4 minus 2x plus 4 upon root over 4x minus x square since x plus 2 is equal to a into 4 minus 2x plus b and a is minus half and b is 4 into dx. So it can further be written as minus half integral 4 minus 2x upon root over 4x minus x square into dx plus integral dx upon root over 4x minus x square. Now let us solve both of these integrals one by one. First we are starting with integral 4 minus 2x upon root over 4x minus x square into dx. Now we are left t is equal to 4x minus x square. So this implies dt is equal to 4 minus 2x into dx. So this integral is further written as dt upon root over t and this is further equal to 2 root over t plus c1 where c1 is a constant so this is equal to 2 root over 4x minus x square plus c1. So this can further be written as minus half into 2 into root over 4x minus x square plus c1. Now we have to integrate this second integral that is dx upon root over 4x minus x square. Now 4x minus x square can be written as 2 square minus x minus 2 whole square. Now using the formula 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 4a square. Now this integral can further be written as dx upon root over 2 square minus x minus 2 whole square. And this is further equal to sin inverse x minus 2 upon 2 plus c2. Thus this integral can be written as 4 into sin inverse x minus 2 upon 2 plus c2. So this is further equal to minus root over 4x minus x square plus 4 times of sin inverse x minus 2 upon 2 plus c1 plus c2 and this is equal to let c. So we have minus root over 4x minus x square plus 4 sin inverse x minus 2 upon 2 plus c. Thus on integrating the given function we get minus root over 4x minus x square plus 4 sin inverse x minus 2 upon 2 plus c. So this completes this equation by integral.