左极限 limf(x) = limx+b = b+1,右极限 limf(x) = limln(x+a) = ln(a+1);则 b+1 = ln(a+1)左导数 lim[f(x)-f(1)]/(x-1) = lim(x-1)/(x-1) = 1,右导数 lim[f(x)-f(1)]/(x-1) = lim[ln(a+x^2)-(b+1)]/(x-1) f(x) 在 x = 1 处可导，则 右导数 = lim2x/(a+x^2) = 2/(a+1) = 1,得 a = 1， b = ln2 - 1,
