Question

1/(x^4-2x^2+1)的不定积分怎么算？？？谢谢

Answer 1

原式=∫dx/(1-x^2)^2,
设x=sint,dx=costdt,
cost=√（1-x^2),tant=x/√（1-x^2),sect=1/√（1-x^2)
原式=∫costdt/(cost)^4
=∫(sect)^3dt
=∫sectd(tant)
=sect*tant-∫tantd(sect)
=sect*tant-∫tant*tant*sectdt
=sect*tant-∫[(sect)^3-sect]dt
∫(sect)^3dt=sect*tant/2+(1/2)∫sectdt
=sect*tant/2+(1/2)ln|sect+tant|+C1
=(1/2)[x/√（1-x^2]/√（1-x^2)+(1/2)ln|1/√（1-x^2)+x/√（1-x^2)|+C1
=(1/2)x/(1-x^2)+(1/2)ln|(x+1)/√(1-x^2)|+C1
=(1/2)x/(1-x^2)+(1/4)ln|(1+x)/(1-x)|+C.

追问

将x换成sint定义域就变了！不过还是谢谢你

Answer 2