Let f(x) be a differentiable function for x > 0 such that f'(x) = x^2 - 1 x^4 + 3x^2 + 1 . If _ x f(x) = 2 , then f(2) is equal to

Step-by-step solution

Given f'(x) = x^2 - 1 x^4 + 3x^2 + 1 . Integrating to find f(x) : f(x) = x^2 - 1 x^4 + 3x^2 + 1 dx Divide the numerator and the denominator by x^2 : f(x) = 1 - 1 x^2 x^2 + 3 + 1 x^2 dx Substitute t = x + 1 x , which gives dt = (1 - 1 x^2 ) dx . The denominator can be written as x^2 + 1 x^2 + 3 = (x + 1 x )^2 - 2 + 3 = t^2 + 1 . The integral becomes: f(x) = 1 t^2 + 1 dt = ⁻¹(t) + C f(x) = ⁻¹ (x + 1 x ) + C Use the given limit condition _ x f(x) = 2 : _ x [ ⁻¹ (x + 1 x ) + C ] = 2 As x , x + 1 x , so ⁻¹( ) = 2 . 2 +