 Hello friends, let's solve the following question. It says integrate the following function. The given function is x upon 9 minus 4 x square. Let's now proceed on with the solution. Let I be the integral x upon 9 minus 4 x square dx. Now here we see that the derivative of 9 minus 4 x square is minus 8 x which contains x. So we put equal to 9 minus 4 x square. So dt by dx is equal to minus 8 x and this implies dt is equal to minus 8 x dx and this implies x dx is equal to minus dt upon 8. So x into dx is minus dt upon 8 and 9 minus 4 x square is t. So substituting all these values integral becomes 1 upon t into minus dt by 8. So this is equal to minus 1 upon 8 into integral of 1 upon t dt. Now we know that the integral of 1 upon x is log x and here we have t. So the integral is minus 1 upon 8 log t plus c and this can be further written as 1 upon 8 log t to the power minus 1 plus c. Now substituting the value of t here, it becomes 1 upon 8 log t is 9 minus 4 x square to the power minus 1 plus c and it is again equal to 1 upon 8 log 1 upon 9 minus 4 x square plus c. Hence the integral of the given function is 1 upon 8 log 1 upon 9 minus 4 x square plus c and this completes the question. Bye for now. Take care. Have a good day.