∫dx/(1+√(1-x^2))
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∫dx/(1+√(1-x^2))∫dx/(1+√(1-x^2))∫dx/(1+√(1-x^2))答：设x=sint原式=∫[1/(1+cost)]d(sint)=∫[cost/(1+cost)]dt=
∫dx/(1+√(1-x^2))
∫dx/(1+√(1-x^2))
∫dx/(1+√(1-x^2))
答：
设x=sint
原式
=∫ [1/(1+cost)]d(sint)
=∫ [cost/(1+cost)]dt
=∫ dt -∫ 1/(1+cost) dt
=t -∫ 1/[cos(t/2)]^2 d(t/2)
=t - tan(t/2)+C
=t- 2sin(t/2)cos(t/2)/2[cos(t/2)]^2+C
=t-sint/(1+cost)+C
=arcsinx-x/[1+√(1-x^2)]+C
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