求解∫1/(1+t)²(1-t)²dt
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/27 15:39:35
![求解∫1/(1+t)²(1-t)²dt](/uploads/image/z/12475930-58-0.jpg?t=%E6%B1%82%E8%A7%A3%E2%88%AB1%2F%EF%BC%881%2Bt%EF%BC%89%26%23178%3B%EF%BC%881-t%EF%BC%89%26%23178%3Bdt)
xAj0CDĆE5n05
#&PW趋.~N3\cJc&zxvaz}PF^
<"6ir
hİA'v~7rk;n0Vu y"ʍmc#%J|U%d;`7A
y$aIxυt>=ez8ΆrW6XɿrqjM]H2gl;`?f]x9
求解∫1/(1+t)²(1-t)²dt
求解∫1/(1+t)²(1-t)²dt
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)