t方减1分之一的不定积分怎么求
问题描述：
t方减1分之一的不定积分怎么求
答
∫1/(t^2-1)dt
=∫1/(t-1)(t+1)dt
=1/2∫[1/(t-1)-1/(t+1)]dt
=1/2∫1/(t-1)d(t-1)-1/2∫1/(t+1)d(t+1)
=1/2ln|t-1|-1/2ln|t+1|+C
=1/2ln|(t-1)/(t+1)|+C
答
令t=secx ∫1/(t^2-1)dt=∫ctanxdx=Log[Sin[x]]+C=Log[Sin[Arccos[1/t]]]+C
答
∫1/(t^2-1)dt
=∫1/(t-1)(t+1)dt
=1/2∫[1/(t-1)-1/(t+1)]dt
=1/2∫1/(t-1)d(t-1)-1/2∫1/(t+1)d(t+1)
=1/2ln|t-1|-1/2ln|t+1|+c