解析：∫1/(4－x²)dx＝－∫1/(x＋2)(x－2)dx＝－1/4∫[1/(x－2)－1/(x＋2)]dx＝－1/4∫1/(x－2)dx＋1/4∫1/(x＋2)dx＝-1/4ln|x－2|＋1/4ln|x＋2|＋C＝1/4ln|(x＋2)/(x－2)|＋C.
原式=(1/3)∫{1/[2^2+(3x)^2]}d(3x)=1/2arctan(3x/2)+C
