1/(t ^2-4）的原函数是？解：∫1/(t²-4)dt=∫1/(t+2)(t-2)dt=∫[1/(t-2)-1/(t+2)]/4dt=[∫d(t-2)/(t-2)-∫d(t+2)/(t+2)]/4=(ln|t-2|-ln|t+2|)/4+C=(ln|(t-2)/(t+2)|)/4+C
好
