∫ (x^2*arccosx)dx=x^3/3*arccosx+[(2+x^2)/9]*根号(1-x^2)+C?

第二项符号似乎不对!
I =∫(x^2*arccosx)dx = (1/3)∫arccosxdx^3
= (1/3)x^3*arccosx + (1/3)∫x^3dx/√(1-x^2),
令 x=sint,则
I1 = ∫x^3dx/√(1-x^2) = ∫(sint)^3dt
= -∫[1-(cost)^2]d(cost) = -cost+(1/3)(cost)^3+3C
= -(1/3)cost[3-(cost)^2]+3C = -(1/3)cost[2+(sint)^2]+3C
= -(1/3)√(1-x^2)(2+x^2)+3C,
则 I = (1/3)x^3*arccosx - (1/9)√(1-x^2)(2+x^2)+C
再问： 书是(1/3)x^3*arccosx - (1/3)∫x^3/√(1-x^2)dx=(1/3)x^3*arccosx -(1/3)∫[x（x^2-1）+x]/√(1-x^2)dx=x^3/3*arccosx+[(2+x^2)/9]*根号(1-x^2)+C
我做：∫x^2*arccosxdx=1/3*x^3arccosx+1/9(1-x^2)^(3/2)-1/3(1-x^2)^(1/2)+C
再答： 你的答案也对。∫x^2*arccosxdx
=1/3*x^3arccosx+1/9(1-x^2)^(3/2)-1/3(1-x^2)^(1/2)+C
=1/3*x^3arccosx-1/9(1-x^2)^(1/2)*(2+x^2)+C。