∫dx/[1+√(1-x^2)] 设x=sint，-π/2<=t<=π/2
=∫[1/(1+cost)]d(sint)
=∫[(1+cost-1)/(1+cost)]dt
=∫[1-1/(1+cost)]dt
=t-∫(1/cos²t/2)d(t/2)
=t+cot(t/2)+C
=arcsinx+[1+√(1-x^2)]/x+C
求不定积分∫1\/(1+根号下1-x平方)
答：∫dx\/[1+√(1-x^2)] 设x=sint，-π\/2<=t<=π\/2 =∫[1\/(1+cost)]d(sint)=∫[(1+cost-1)\/(1+cost)]dt =∫[1-1\/(1+cost)]dt =t-∫(1\/cos²t\/2)d(t\/2)=t+cot(t\/2)+C =arcsinx+[1+√(1-x^2)]\/x+C ...
不定积分1\/(1+根号下1—x^2)
=∫(1-√(1-x^2))\/x^2dx x=sint，√(1-x^2)=cost，dx=costdt，t∈[0,π\/2]=∫(1-cost)\/sin²t*costdt =∫[cost\/sin²t-cos²t\/sin²t]dt =-∫dsint\/sin²t-∫cot²tdt =1\/sint-∫(csc²t-1)dt =1\/sint+∫(-csc²t...
求不定积分∫1\/1+根号(1-x^2)dx
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
1\/(1+根号里面(1-x的平方)) 的不定积分 求详细过程
所以，原式=arcsinx+{[√(1-x^2)]\/x}+(1\/x)+C
1\/[1+根号(1-x^2)] 该式的不定积分为什么? \/表示分数线, x^2表示x...
x=sint -兀\\2
求∫[1+根号下1-x的平方]的-1次方的不定积分
先分母有理化：∫1\/(1+√(1-x^2)dx=∫(1-√(1-x^2)\/x^2dx=∫dx\/x^2-∫√(1-x^2)\/x^2dx对第二个积分,由分部积分：∫√(1-x^2)\/x^2dx=∫√(1-x^2)d(1\/x)=√(1-x^2)\/x-∫(1\/x)d√(1-x^2)=√(1-x^2)\/xarcsinx+C所以：∫1...
用换元法求不定积分 ∫ dx\/1+根号(1-X^2)
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求∫[1+根号下1-x的平方]的-1次方的不定积分
∫1\/(1+√(1-x^2)dx=∫(1-√(1-x^2)\/x^2dx=∫dx\/x^2-∫√(1-x^2)\/x^2dx对第二个积分,由分部积分:∫√(1-x^2)\/x^2dx=∫√(1-x^2)d(1\/x)=√(1-x^2)\/x-∫(1\/x)d√(1-x^2)=√(1-x^2)\/xarcsinx+C所以:∫1\/(1+√(1-x^2)dx=-1\/x+√(1-x^2)\/x+arcsinx...
[1+根号下(1-x的平方)]分之一的不定积分
令x=sint -π\\2<=t<=π\\2 则dx=costdt ∫1\\[1+(1-x^2)^(1\\2)]dx =∫costdt\\(1+cost)=∫[1-1\\(1+cost)]dt =∫{1-(1\\2)*[sec(t\\2)]^2}dt =t-tan(t\\2)+C =arcsinx-x\\[1+(1-x^2)^(1\\2)]+C ...
