∫ (x+1)/(根号下1-x^2)dx

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∫粗薯 (x+1)/岩孝者√(1-x^2) dx
let
x=siny
dx = cosy dy

∫ (x+1)/√慎丛(1-x^2) dx
=∫ (siny+1) dy
=-cosy + y + C
=-√(1-x^2) + arcsinx + C

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