 Hello and welcome to the session. The question says integrate the function second is root over x square plus 3 x. So firstly let us learn the formula to integrate the function of the type x square minus a square with respect to x. So its formula is x upon 2 into root over x square minus a square minus a square upon 2 into log mod x plus root over x square minus a square plus c where c is a constant. So this formula is the key idea to the question with the help of which we shall be solving the problem. Let us now start with the solution. The given function is root over x square plus 3 x. Let us now consider the polynomial x square plus 3 x. Now we shall try to write this polynomial as the sum of the squares of two polynomials with the help of this. But if we have a polynomial x square plus b x plus c then it can be written as a into x plus b upon 2 into a whole square plus c upon a minus b square upon 4 a square. So here a is 1, b is 3 and c is 0. So x square plus 3 x can be written as 1 into x plus 3 upon 2 into 1 whole square plus c is 0. So we have 0 upon 1 minus b square upon 4 a square. So b is 3 so we have b square as 9 minus 4 into 1 whole square. So this is further equal to x plus 3 upon 2 whole square plus into minus is minus 3 upon 2 whole square. Thus the given function which is root over x square plus 3 x can be written as root over x plus 3 upon 2 whole square minus 3 upon 2 whole square which is in the form of x square minus a square. So we have to integrate the function root over x plus 3 upon 2 whole square minus 3 upon 2 whole square with respect to x. Now let us take t is equal to x plus 3 upon 2. So this implies dt is equal to dx by differentiating both sides with respect to x. This function can further be written as root over t square minus 3 upon 2 whole square into dt. Now let us apply the formula to integrate of the function of the type x square minus a square here x is t and a is 3 upon 2. So it is formalize t upon 2 into root over t square minus 3 upon 2 whole square minus a square upon 2 so a is 3 upon 2 so 3 upon 2 whole square upon 2 log mod t plus root over t square minus 3 upon 2 whole square plus c where c is a constant. It can further be written as t upon 2 and t is x plus 3 upon 2 that is 2 x plus 3 upon 2 and we have 2 in the denominator into root over t square minus 3 upon 2 whole square is x plus 3 upon 2 whole square minus 3 upon 2 whole square and this is equal to x square plus 3 x. So here we have x square plus 3 x minus on simplifying this we get 9 upon 8 log into x plus 3 upon 2 plus the value of this root is x square plus 3 x root over right that is on integrating the given function we get plus 3 upon 4 into root over x square plus 3 x minus 9 upon 8 log mod x plus 3 upon 2 t can be written as 2 x plus 3 upon 2 whole square plus root over x square plus 3 x plus a constant c so this complete the session. Bye and take care.