The value of _ x 4 1 (x - 4 )^2 _ 4 ^ x ( 2 ^2 t + 3 t + 4 - ^2 t + 2 t + 6 ) dt is equal to

Step-by-step solution

The given limit is of the form 0 0 . We can apply L'Hôpital's rule and use the Newton-Leibniz formula to differentiate the integral. _ x 4 d dx _ 4 ^ x ( 2 ^2 t + 3 t + 4 - ^2 t + 2 t + 6 ) dt d dx (x - 4 )^2 _ x 4 2 ^2 x + 3 x + 4 - ^2 x + 2 x + 6 2 (x - 4 ) Rationalizing the numerator: _ x 4 2 ^2 x + 3 x + 4 - ( ^2 x + 2 x + 6) 2 (x - 4 ) ( 2 ^2 x + 3 x + 4 + ^2 x + 2 x + 6 ) _ x 4 ^2 x + x - 2 2 (x - 4 ) ( 9 + 9 ) _ x 4 ( x - 1)( x + 2) 12 (x - 4 ) We know that _ x 4 x - 1 x - 4 is the derivative of x at x = 4 ,