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∫xArCtAnxDx

∫xarctanxdx =(1/2)∫ arctanxd(x²) 那么使用分部积分法得到, =(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+x&#...

分部积分思想: ∫x^2arctanxdx=(1/3)∫arctanxdx^3 =(1/3)x^3arctanx-(1/3)∫x^3darctanx =(1/3)x^3arctanx-(1/3)∫[(x^3+x)-x]/(1+x^2)dx =(1/3)x^3arctanx-(1/3)∫xdx+(1/3)∫(x)/(1+x^2)dx =(1/3)x^3arctanx-(1/6)x^2+(1/6)ln(1+x^2)+C(C为常数)...

因为这里书写不便,故将我的答案做成图像贴于下方,谨供楼主参考。 (若图像显示过小,点击图片可放大)

∫ x²arctanx dx = ∫ arctanx d(x³/3) = (1/3)x³arctanx - (1/3)∫ x³ d(arctanx) = (1/3)x³arctanx - (1/3)∫ x³/(1 + x²) dx = (1/3)x³arctanx - (1/3)∫ x[(1 + x²) - 1]/(1 + x²) dx = (1/3)...

用分部积分法, ∫ x² arctanx dx =1/3 ∫ arctanx d(x³) =1/3 x³ arctanx - 1/3 ∫x³/(1+x²) dx =.......后面会了吧

∫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)x...

答案在图片上,满意请点采纳 千万别点错哦,那会上当的,解题用不着压缩包 所以那些人上传的压缩包可能是含有病毒的 愿您学业进步,谢谢☆⌒_⌒☆

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