∫x^2/√(1-x^2)dx 的不定积分

∫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=(...