 Hello and welcome to the session my name is Asha and I am going to help you with the following question it says integrate the following function 7th is root over 1 plus 3x minus x square. So let us start with the solution and we have to integrate root over 1 plus 3x minus x square with respect to x. We have a polynomial of the type ax square plus bx plus c and it can also be written as a into x plus b upon 2a whole square plus c upon a minus b square upon 4a square. So let us try to write 1 plus 3x minus x square as the sum of the squares of two polynomials with the help of this. So first it can be written as minus of x square minus 3x minus 1 it is further equal to minus x minus 3 upon 2 whole square plus c upon a c is minus 1 upon 1 minus b square upon 4a square. So b is minus 3 so we have 9 upon 4 into 1 square it is further equal to minus x minus 3 upon 2 whole square minus on simplifying this here we get 13 upon 4 which can further be written as root over 13 upon 2 whole square by opening this bracket minus x minus 3 upon 2 whole square. So this integral can further be written as root over root over 13 upon 2 whole square minus x minus 3 upon 2 whole square into dx. Now let us take x is equal to x minus 3 upon 2 this implies dt is equal to dx so this integral can further be written as integral root over root over 13 upon 2 whole square minus t square into dt. Now the formula to integrate the function of the type a square minus x square with respect to x is equal to half x into root over a square minus x square plus a square upon 2 sin inverse x upon a plus c. So with the help of this integral formula this integral can further be written as x upon 2 that is t upon 2 root over a square minus x square is root over 13 upon 2 and x is t. So here we can write as root over 13 upon 2 whole square minus t square plus a square upon 2 a is root over 13 upon 2 whole square upon 2 sin inverse x upon a that is t upon root over 13 upon 2 plus c. This is further equal to dx x minus 3 upon 2 upon 2 and the value of this root which is root over 13 upon 2 whole square minus t square is 1 plus 3x minus x square since from here only we derived this and from this we substituted dx x minus 3 upon 2. So here we can write 1 plus 3x minus x square then on solving this we have 13 upon 8 sin inverse dx x minus 3 upon 2 upon root over 13 upon 2 plus c. This can further be written as 2x minus 3 upon 4 into root over 1 plus 3x minus x square plus 13 upon 8 sin inverse 2x minus 3 upon root over 13 plus c. Thus when integrating the given function we get x minus 3 upon 4 into root over 1 plus 3x minus x square plus 13 upon 8 sin inverse 2x minus 3 upon root over 13 plus a constant c. So this completes the session. Bye and take care.