Respuesta :

Delaney1217
So, f[x] = 1/4x^2 - 1/2Ln(x) 
thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x 
thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) 
thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 
thus Sqrt[...] = (x/2 + 1/2x)