问题标题：
(x的平方)乘以arccotx的不定积分怎么求?$(acontent)
问题描述：
(x的平方)乘以arccotx的不定积分怎么求?
$(acontent)
方叶祥回答：
∫x^2(arccotx)dx
=(1/3)∫(arccotx)dx^3
=(1/3)x^3arccotx-(1/3)∫x^3darccotxdarccotx=darctan(1/x)=(-1/x^2)[1/(1+1/x^2)]=-1/(1+x^2)
=(1/3)x^3arccotx+(1/3)∫x^3dx/(1+x^2)
=(1/3)x^3arccotx+(1/12)∫dx^4/(1+x^2)
=(1/3)x^3arccotx+(1/6)∫x^2dx^2/(1+x^2)
=(1/3)x^3arccotx+(1/6)∫dx^2-(1/6)∫dx^2/(1+x^2)
=(1/3)x^3arccotx+(1/6)x^2-(1/6)ln(1+x^2)+C
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