高数求不定积分，4，5，6求详细过程呀ⷶⷶ

4.∫dx/(1+√(1-x^2))
x=sinu dx=cosudu √(1-x^2)=cosu
tan(u/2)=sinu/(1+cosu)=x/(1+√(1-x^2))
=∫cosudu/(1+cosu)
=∫[1-1/(1+cosu)]du
=u-∫du/(1+cosu)
=u-∫d(u/2)/(cos(u/2))^2
=u-tan(u/2)+C
=arcsinx - x/(1+√(1-x^2)) +C
5.设x=tanb
则原函数=∫1/√(x^2+1)^3dx
=∫1/√(tanb^2+1)^3dtanb
=∫-cosb db
= - sinb + c
= - x/√（x^2+1）+c
6.令x=secz，dx=secztanz dz，(暂时不讨论角度z的范围)
√(x²-1)=√(sec²z-1)=√tan²z=tanz
tanz=√(x²-1)/1
sinz=√(x²-1)/√[1²+√(x²-1)²]=√(x²-1) / x
∫1/[x²√(x²-1)] dx
=∫1/(sec²z*tanz) * secztanz dz
=∫1/secz dz
=∫cosz dz
=sinz + C
=√(x²-1) / x + C