 Hello and welcome to the session. The question says, integrate the following function and the given function is 1 upon root over 7 minus 6x minus x square. So, first let us learn to integrate the function of the type a square minus x square with respect to x. So, we are going to sign inverse x upon a plus c where c is a constant and if we have a polynomial of the type ax square plus bx plus c, then to write this polynomial as the sum of squares of two polynomials, we use the formula a into x plus b upon 2a whole square plus c upon a minus b square upon 4a square. This you can also see on page number 614 of your book, here to solve integral of these type, we first reduce these polynomials as the sum of the squares of two polynomials. So, with the help of these two ideas, we shall be solving the above problem. So, this is our key idea. Now, let us start with the solution. First let us consider 7 minus 6x minus x square, this can be written as minus of x square plus 6x minus 7 or it can further be written as minus by applying this formula, we have x plus bx 6 upon 2 into 1 whole square plus c upon a, here we have c as minus 7 upon a is 1 minus b square upon 4a square. So, b square is 36 into upon 4 into 1 square. So, this is further equal to minus x plus 3 whole square plus minus 7 minus 4 9s of 36. So, we have 9 bracket close and this is further equal to minus 7 minus 9 gives minus 16. So, we have 16 minus x plus 3 whole square or we have 4 square minus x plus 3 whole square. So, the given function with us 1 upon root over 7 minus 6x minus x square can be written as 1 upon root over 4 square minus x plus 3 whole square. Now, we have to integrate this function with respect to x, that is we have to find 1 upon root over 4 square minus x plus 3 whole square into dx. First let us put t is equal to x plus 3. So, this implies dt is equal to dx. So, this integral can further be written as dt upon root over 4 square minus t square. Now, by applying the key idea, now that is if we have an integral of the type 1 upon root over a square minus x square, then its integration is equal to sin inverse x upon a plus c. So, this can be written as sin inverse t upon 4 plus c where c is a constant. Let us now put the value of t which is x plus 3. So, we have sin inverse x plus 3 upon 4 plus c. That is when integrating the given function, we get sin inverse x plus 3 upon 4 plus c. So, this completes the session. Bye and take care.