I already asked this question once today but I guess Im set

I already asked this question once today, but I guess I\'m set in my ways, b/c I didn\'t agree with the two previous responses. Is my work incorrect? If so, please explain

Solution

f(x) = xarcTan(2x) - 1/4 ln (1+4x^2) f\'(x) = arctan(2x) + x * 1/ 1+(2x)^2 *2 - 1/4 * 1/ (1+4x^2) * 8x = arctan(2x) +2x/(1+4x^2) -2x/1+4x^2 = arctan(2x)