 Hello and welcome to the session the question says integrate the following function and the 14th one is 1 upon 9x square plus 6x plus 5. So first let us learn how do we change a polynomial of the form ax square plus bx plus c as the sum of squares of two polynomials. So it is formula is a into x plus b upon 2a whole square plus c upon a minus b square upon 4a square. So to find the integral we just have the form dx upon ax square plus bx plus c we first change the denominator as the sum of the squares of two polynomials. See page number 614 of your book to learn more about this. So this is the key idea we are going to use in this problem. Let us now start with the solution. Now here first let us consider the denominator which is 9x square plus 6x plus 5 and let us try to write it as the sum of squares of two polynomials. So first we have 9 into x plus b upon 2a so b is 6 2 into a is 2 into 9 that is 18 whole square minus sorry plus c upon a c is 5 and a is 9 minus b square upon 4a square. So b is 6 so b square is 36 minus 4 into a square and a is 9 so a square is 81. This is further equal to 9 into x plus 1 upon 3 whole square since 6 3s are 18 plus 5 upon 9 minus 4 9s are 36 and 9 9s are 81 so we have 1 upon 9. This further implies 9 into x plus 1 upon 3 whole square plus 5 upon 9 minus 1 upon 9 is 4 upon 9 which can be written as 2 upon 3 whole square thus the given function which is 1 upon 9x square plus 6x plus 5 can be written as 1 upon 9 into x plus 1 upon 3 whole square plus 2 upon 3 whole square. Now we have to integrate this function that is we have to find integral 1 upon 9 into x plus 1 upon 3 whole square plus 2 upon 3 whole square into dx. Firstly let us put x plus 1 upon 3 is equal to t so this implies dx is equal to dt on differentiating both sides with respect to x so it can further be written as taking the constant outside the integral sign dt upon t square plus 2 upon 3 whole square. Now to integrate the function of the type 1 upon x square plus a square with respect to x this is equal to 1 upon a into tan inverse x upon a plus c where c is the constant thus this can be written as 1 upon 9 into now differentiating this function sorry integrating this function we have 1 upon 2 upon 3 into tan inverse x upon a that is t upon 2 upon 3 plus c which is further equal to 1 upon 9 into 3 upon 2 tan inverse t is 3x plus 1 upon 3 into taking this denominator plus c now this is further equal to 3 into 3 is 9 so we have 1 upon 6 tan inverse on simplifying we have 3x plus 1 upon 2 plus c thus on integrating the given function we get 1 upon 6 tan inverse 3x plus 1 upon 2 plus c this completes the session by intake care