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∫x^2/√(1-x^2)dx 的不定积分
人气:186 ℃ 时间:2020-02-05 14:25:15
解答
∫x^2/√(1-x^2)dx=-∫ -2x^2/2√(1-x^2) dx=-∫ x d√(1-x^2)=-x√(1-x^2)+∫√(1-x^2)dx其中,解∫√(1-x^2)dx令x=sintdx=costdt则∫√(1-x^2)dx= ∫(cost)^2 dt=(1/2)∫(1+cos2t) dt=(1/2)cost+(1/4)∫cos2t d2t=(...
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