If ₀^ k 3 - x^2 (3 + x^2) 9 + x^4 dx = 6 6 and k > 3 , then the greatest integer less than or equal to k is ______.

Step-by-step solution

Let I = ₀^ k 3 - x^2 (3 + x^2) 9 + x^4 dx . Dividing the numerator and denominator by x^2 : I = ₀^ k 3 x^2 - 1 (x + 3 x ) x^2 + 9 x^2 dx I = ₀^ k 3 x^2 - 1 (x + 3 x ) (x + 3 x )^2 - 6 dx Let t = x + 3 x . Then dt = (1 - 3 x^2 ) dx ( 3 x^2 - 1 ) dx = -dt . When x 0 , t . When x = k , t = k + 3 k . I = _ ^ k + 3 k -dt t t^2 - 6 = _ k + 3 k ^ dt t t^2 - 6 I = [ 1 6 ⁻¹ ( t 6 ) ]_ k + 3 k ^ I = 1 6 ( 2 - ⁻¹ ( k^2 + 3 k 6 ) ) = 1 6 ⁻¹ ( k^2 + 3 k 6 ) Given I = 6 6 , we have: 1 6 ⁻¹ ( k^2 + 3 k 6 ) = 6 6 ⁻¹ ( k^2 + 3 k 6