-2∫(1 / (t^2-1) )dt 积分的运算与转换问题.
4个回答

先将被积函数化为两个简单函数之和,然后分项积分.

原式=-2∫1/[(t-1)(t+1)]dt(下面裂项)

=-∫{[1/(t-1)]-[1/(t+1)]}dt(下面分项再用凑微分法)

=-{∫[1/(t-1)]d(t-1)-∫[1/(t+1)]d(t+1)}

=-[ln|t-1|-ln|t+1|]+C

=-ln|(t-1)/(t+1)|+C

That' s all.