lim(x→0) arctanx / 2x ＝1/2怎么来的

分部积分法
∫arctanx/(x-1)^3dx
=-2∫arctanxd1/(x-1)^2
=-2(arctanx*1/(x-1)^2-∫1/(x-1)^2darctanx)
=-2(arctanx*1/(x-1)^2-∫1/(x-1)^2*1/(1+x^2)dx)
∫1/(x-1)^2*1/(1+x^2)dx
=∫1/[(x-1)^2*(1+x^2)]dx
已经到达有理函数积分了
设1/(x-1)^2*1/(1+x^2)=A/(x-1)+B/(x-1)^2+(Cx+D)/(x^2+1)
分别解得A=-1/2 B=1/2 C=1/2 D=0
∫1/(x-1)^2*1/(1+x^2)dx
=-1/2∫1/(x-1)-1/(x-1)^2-x/(x^2+1)dx
=-1/2[∫1/(x-1)dx-∫1/(x-1)^2dx-∫x/(x^2+1)dx]
=-1/2[ln｜x-1｜+1/(x-1)-1/2ln(x^2+1)]+C
所以
∫arctanx/(x-1)^3dx
=-2(arctanx*1/(x-1)^2+1/2[ln｜x-1｜+1/(x-1)-1/2ln(x^2+1)]+C
ok.