Question

怎么用第二类换元法求，∫1/（X∧4-1）？

Answer 1

现将原式进行变换
原式=∫dx/[(x^2-1)(x^2+1)]
=∫1/2dx*[1/(x^2-1)-1/(x^2+1)]
=1/2∫dx[/(x^2-1)-1/(x^2+1)]
=1/2[∫dx/(x^2-1)-∫dx/(x^2+1)]

我们分开算，先算∫dx/(x^2-1)部分：
x=sin(t),dx=cos(t)dt 带入,求得积分
∫dx/(x^2-1)=∫cos(t)dt/(-cos(t))^2
=-∫cos(t)dt/cos(t)^2
=-∫dsin(t)/[(1-sin(t))(1+sin(t))]
=-(1/2)ln(|1+sin(t)|/|1-sin(t)|)+C
=-(1/2)ln|(1+x)/(1-x)|+C
再算∫dx/(x^2+1)部分
令x=tan(t),dx=(sect)^2 dt带入,求得积分∫dx/(1+x^2)=∫[sec(t)]^(-2)d(tant)=∫dt=t+c=arctanx+C

原式=1/2*[-(1/2)ln|(1+x)/(1-x)|-arctanx]+C