∫xarctanxdx 求详细过程

Question

如题所述

Answer 1

∫xarctanxdx
=(1/2)∫ arctanxd(x²)
分部积分
=(1/2)x²arctanx - (1/2)∫ x²/(1+x²) dx
=(1/2)x²arctanx - (1/2)∫ (x²+1-1)/(1+x²) dx
=(1/2)x²arctanx - (1/2)∫ 1 dx + (1/2)∫ 1/(1+x²) dx
=(1/2)x²arctanx - (1/2)x + (1/2)arctanx + C

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Answer 2

这个得用分部积分……
∫xarctanxdx= (1/2)x²arctanx-∫ (1/2)x²(arctanx)'dx
= (1/2)x²arctanx-(1/2)∫ x²/(1+x²)dx
=1/2)x²arctanx - (1/2)∫ (x²+1-1)/(1+x²) dx
=(1/2)x²arctanx - (1/2)∫ 1 dx + (1/2)∫ 1/(1+x²) dx
=(1/2)x²arctanx - (1/2)x + (1/2)arctanx + C
其中原函数是(1/2)x²arctanx
v=(1/2)x² v'=x u=arctanx u'=1/(1+x²)

Answer 3

解答：

∫xarctanxdx=(1/2)x^2arctanx-(1/2)∫x^2/(1+x^2)dx=(1/2)x^2arctanx-(1/2)∫[1-1/(1+x^2)]dx

=(1/2)x^2arctanx-(1/2)x^2-(1/2)arctanx+C