求解∫1/（1+t）²（1-t）²dt

∫dt /[ (1+t)^2.(1-t)^2 ]
= ∫dt / (1-t^2)^2
let
t= sinx
dt = cosx dx
∫dt / (1-t^2)^2
=∫ dx/ (cosx)^3
=∫ (secx)^3dx
consider
∫ (secx)^3dx = ∫ secx dtanx
= secx tanx - ∫ secx (tanx)^2 dx
= secx tanx - ∫ [(secx)^3-secx ]dx
2∫ (secx)^3dx =secx tanx + ∫ secxdx
=secx tanx + ln|secx+tanx|
∫ (secx)^3dx = (1/2)[secx tanx + ln|secx+tanx|] +C
ie
∫dt / (1-t^2)^2
=∫ (secx)^3dx
=(1/2)[secx tanx + ln|secx+tanx|] +C
=(1/2)[ t/(1-t^2) + ln|1/√(1-t^2) + t/√(1-t^2)| ] + C
where
t=sinx
cosx =√(1-t^2)