Question

对1/(1-t^2)^2求原函数

Answer 1

∫dt/(1-t^2)^2
let
t= siny
dt =cosydy
∫dt/(1-t^2)^2
=∫dy/(cosy)^3
=∫(secy)^3dy
consider
∫(secy)^3dy = ∫secydtany
=secy.tany - ∫secy.(tany)^2dy
=secy.tany - ∫secy.[(secy)^2-1]dy
2∫(secy)^3dy =secy.tany + ∫secydy
=secy.tany + ln|secy+tany|
∫(secy)^3dy = (1/2)[secy.tany + ln|secy+tany| ] + C
=(1/2)[secy.tany + ln|secy+tany| ] + C
= (1/2)[ t/(1-t^2) + ln|√(1-t^2) + t/√(1-t^2) ] + C
∫dt/(1-t^2)^2
=∫(secy)^3dy
=(1/2)[ t/(1-t^2) + ln|√(1-t^2) + t/√(1-t^2) ] + C