dt/（1+根号（1+t）） 求不定积分
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∫dt /(1+√(1+t) )
let
t = (tany)^2
dt = 2tany (secy)^2 dy
∫dt /(1+√(1+t) )
=∫2tany (secy)^2 /(1+secy) dy
=2∫ siny/[(cosy)^2(1+cosy)] dy
let
z= cosy
dz = siny dy
2∫ siny/[(cosy)^2(1+cosy)] dy
=2∫ dz/[z^2(1+z)]
let
1/[z^2(1+z)] = a1/z+ a2/z^2+ b/(1+z)
=>
1= a1z(1+z) + a2(1+z) +bz^2
z=0, a2=1
z=-1, b=1
coef. of z^2
a1+b=0
a1=-1
1/[z^2(1+z)] =-1/z+ 1/z^2+ 1/(1+z)
2∫ dz/[z^2(1+z)]
=2∫ [-1/z+ 1/z^2+ 1/(1+z)]dz
=2(ln|(1+z)/z| - 1/z ) + C
where z = 1/√(1+t)
t = (tany)^2
tany = √t
z= cosy =1/√(1+t)
let
t = (tany)^2
dt = 2tany (secy)^2 dy
∫dt /(1+√(1+t) )
=∫2tany (secy)^2 /(1+secy) dy
=2∫ siny/[(cosy)^2(1+cosy)] dy
let
z= cosy
dz = siny dy
2∫ siny/[(cosy)^2(1+cosy)] dy
=2∫ dz/[z^2(1+z)]
let
1/[z^2(1+z)] = a1/z+ a2/z^2+ b/(1+z)
=>
1= a1z(1+z) + a2(1+z) +bz^2
z=0, a2=1
z=-1, b=1
coef. of z^2
a1+b=0
a1=-1
1/[z^2(1+z)] =-1/z+ 1/z^2+ 1/(1+z)
2∫ dz/[z^2(1+z)]
=2∫ [-1/z+ 1/z^2+ 1/(1+z)]dz
=2(ln|(1+z)/z| - 1/z ) + C
where z = 1/√(1+t)
t = (tany)^2
tany = √t
z= cosy =1/√(1+t)
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