Question

不定积分

1/(根号(1-u^2)+u)

Answer 1

∫du/[√(1-u²)+u]，令u=sinx
=∫cosx/(cosx+sinx) dx，上下除以cosx
=∫dx/(1+tanx)，令y=tanx
=∫dy/[(1+y)(1+y²)]
令1/[(1+y)(1+y²)]=(A+By)/(1+y²)+C/(1+y)
A=1/2，B=-1/2，C=1/2
=(1/2)∫dy/(1+y²)-(1/2)∫ydy/(1+y²)+(1/2)∫dy/(1+y)
=(1/2)arctany-(1/4)ln|1+y²|+(1/2)ln|1+y| + C
=(1/2)arctan(tanx)-(1/4)ln|sec²x|+(1/2)ln|1+tanx| + C
=x/2+(1/2)ln|sinx+cosx| + C
=(1/2)arcsin(u)+(1/2)ln|u+√(1-u²)| + C