发布网友 发布时间:2024-05-08 12:03
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热心网友 时间:2024-05-12 17:58
∫x(arctanx)²dx=1/2∫(arctanx)²dx²=1/2(xarctanx)²-∫(x²arctanx)/(1+x²)dx=1/2(xarctanx)²-∫[arctanx-arctanx/(1+x²)]dx=1/2(xarctanx)²-∫arctanxdx-∫arctanxd(arctanx)=1/2(xarctanx)²-xarctanx+∫x/(1+x²)dx-1/2(arctanx)²=1/2(xarctanx)²-xarctanx+1/2ln(1+x²)-1/2(arctanx)²+C