 Hello and welcome to the session. The question says, integrate the following function 18th is 1 upon root over 8 plus 3x minus x square. So first let us learn, if we have a polynomial of the type Ax square plus Bx plus c, then to change it as the sum of the squares of 2 polynomials, it is formalized A into x plus B upon 2a whole square plus c upon A minus B square upon 4a square, which is on page number 614 of your book to find the integral of the type whose denominator is a polynomial. So this is the P idea we are going to use to integrate the given function. Let us now start with the solution. Firstly, let us consider the polynomial 8 plus 3x minus x square which can be written as minus of x square minus 3x minus 8. This is further equal to minus x plus minus 3 upon 2 into 1 that is 2 whole square plus c upon A that is minus 8 upon 1 minus B square upon 4a square. So B is minus 3, so B square is 9 upon 4 into 1 square. So this is further equal to minus x minus 3 upon 2 whole square plus into minus s minus and on simplifying this we have 8 plus 32 and 32 plus minus 41. So we have 41 upon 4 which can further be written as root over 41 upon 2 whole square minus x minus 3 upon 2 whole square. That is the given function can be written as 1 upon root over root over 41 upon 2 whole square minus x minus 3 upon 2 whole square. Now we have to integrate this function, so we have to find the value of this integral. This is further equal to, first let us put x minus 3 upon 2 is equal to t. So in differentiating both sides it is 2x we have dx is equal to dt integral further implies dt upon root over root over 41 upon 2 whole square minus t square and to solve the integral of the type 1 upon root over a square minus x square with respect to x its formula is sin inverse x upon a plus c. Thus this integral can further be written as sin inverse t upon root over 41 upon 2 plus c or this is further equal to sin inverse t s 2x minus 3 upon 2 into 2 upon root over 41 plus c taking this to enumerator. This further implies that sin inverse 2 cancels out with 2 and we have 2x minus 3 upon root over 41 plus c. Thus when integrating the given function our answer is sin inverse 2x minus 3 upon root over 41 plus c. This completes the session. Bye and take care.