Let f(x) be a differentiable function for x > 0 such that f'(x) = 12x^7 + 6x^5 (2x^6 + 3x^2 + 1)^2 . If f(1) = 2 3 , then the value of _ x f(x) is equal to

Step-by-step solution

Given f'(x) = 12x^7 + 6x^5 (2x^6 + 3x^2 + 1)^2 Dividing the numerator and the denominator by x¹² , we get: f'(x) = 12x⁻⁵ + 6x⁻⁷ (2 + 3x⁻⁴ + x⁻⁶)^2 Integrating both sides with respect to x : f(x) = 12x⁻⁵ + 6x⁻⁷ (2 + 3x⁻⁴ + x⁻⁶)^2 dx Let 2 + 3x⁻⁴ + x⁻⁶ = t Differentiating both sides, we get: (-12x⁻⁵ - 6x⁻⁷) dx = dt (12x⁻⁵ + 6x⁻⁷) dx = -dt Substituting this into the integral: f(x) = -dt t^2 = 1 t + C f(x) = 1 2 + 3x⁻⁴ + x⁻⁶ + C = x^6 2x^6 + 3x^2 + 1 + C Given that f(1) = 2 3 : f(1) = 1 2 + 3 + 1 + C = 1 6 + C 1 6 + C