求1/(1+x^2)(1-x^2)^0.5的不定积分
答案:1 悬赏:20
解决时间 2021-01-16 12:17
- 提问者网友：我一贱你就笑
- 2021-01-15 23:33
求1/(1+x^2)(1-x^2)^0.5的不定积分
最佳答案
- 二级知识专家网友：山君与见山
- 2021-01-16 00:02
Let t = √[(1-x??)/(1+x??)
t?? = (1-x??)/(1+x??)
2tdt = [(1+x??)(-2x)-(1-x??)(2x)]/(1+x??)?? dx = -4x/(1+x??)?? dx => tdt = -2xdx/(1+x??)??
x?? = (1-t??)/(1+t??)，1+x?? = 2/(1+t??)，1-x?? = 2t??/(1+t??)
∫ dx/[(1+x??)√(1-x??)] = ∫ (1+x??)/[-2x√(1-x??)] * [-2xdx/(1+x??)??]
= ∫ [2/(1+t??)] / {[-2√(1-t??)/(1+t??)] * √[2t??/(1+t??)]} * tdt
= ∫ [2/(1+t??)] / [-2√(1-t??)/√(1+t??) * √2*t/√(1+t??)] * tdt
= -(1/√2)∫ 1/√(1-t??) dt
= -(1/√2)arcsin(t) + C
= -(1/√2)arcsin√[(1-x??)/(1+x??)] + C
t?? = (1-x??)/(1+x??)
2tdt = [(1+x??)(-2x)-(1-x??)(2x)]/(1+x??)?? dx = -4x/(1+x??)?? dx => tdt = -2xdx/(1+x??)??
x?? = (1-t??)/(1+t??)，1+x?? = 2/(1+t??)，1-x?? = 2t??/(1+t??)
∫ dx/[(1+x??)√(1-x??)] = ∫ (1+x??)/[-2x√(1-x??)] * [-2xdx/(1+x??)??]
= ∫ [2/(1+t??)] / {[-2√(1-t??)/(1+t??)] * √[2t??/(1+t??)]} * tdt
= ∫ [2/(1+t??)] / [-2√(1-t??)/√(1+t??) * √2*t/√(1+t??)] * tdt
= -(1/√2)∫ 1/√(1-t??) dt
= -(1/√2)arcsin(t) + C
= -(1/√2)arcsin√[(1-x??)/(1+x??)] + C
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