∫xarctanxdx
=1/2∫arctanx*2xdx
=1/2∫arctanxdx^2
=1/2xarctanx-1/2∫x^2*1/(x^2+1)dx
=1/2xarctanx-1/2∫(x^2+1-1)dx/(x^2+1)
=1/2xarctanx-1/2∫dx+1/2∫dx/(x^2+1)
=1/2xarctanx-x/2+1/2*arctanx+C
=1/2*(xarctanx-x+arctanx)+C
黎曼积分
定积分的正式名称是黎曼积分，用黎曼自己的话来说，就是把直角坐标系上的函数的图象用平行于y轴的直线把其分割成无数个矩形，然后把某个区间[a,b]上的矩形累加起来。
所得到的就是这个函数的图象在区间[a,b]的面积，实际上，定积分的上下限就是区间的两个端点a,b。
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第1个回答 推荐于2019-01-14
∫xarctanxdx
=∫arctanxdx²/2
=x²/2arctanx-∫x²/2darctanx
=x²/2arctanx-1/2∫x²/(1+x²)dx
=x²/2arctanx-1/2∫(x²+1-1)/(1+x²)dx
==x²/2arctanx-1/2∫1-1/(1+x²)dx
==x²/2arctanx-1/2x+1/2arctanx+c本回答被网友采纳
=∫arctanxdx²/2
=x²/2arctanx-∫x²/2darctanx
=x²/2arctanx-1/2∫x²/(1+x²)dx
=x²/2arctanx-1/2∫(x²+1-1)/(1+x²)dx
==x²/2arctanx-1/2∫1-1/(1+x²)dx
==x²/2arctanx-1/2x+1/2arctanx+c本回答被网友采纳
第2个回答 2012-11-15
∫xarctanxdx
= (1/2)∫ arctanx dx^2
= (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 - x/2 + (1/2)∫ 1/(1+x^2) dx
=(1/2)x^2arctanx - x/2 + (1/2) arctanx + C
= (1/2)∫ arctanx dx^2
= (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 - x/2 + (1/2)∫ 1/(1+x^2) dx
=(1/2)x^2arctanx - x/2 + (1/2) arctanx + C
第3个回答 2012-11-15
∫xarctanxdx =1/2x^2*arctanx-∫1/2x^2darctanx= 1/2x^2*arctanx-∫1/2*x^2*1/(1+x^2)dx=1/2x^2*arctanx-1/2∫x^2*1/(1+x^2)dx=1/2x^2*arctanx-1/2∫(1-1/(1+x^2))dx=/2x^2*arctanx-1/2*x+1/2*arctanx+c
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