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∫xln x 1 Dx

∫xln(x-1)dx 的不定积分是多少?=x^2/2* ln(x-1)-x^2/4-∫(x-1)/2(x-1)dx-∫1/2(x-1)dx =x^2/2* ln(x-1)-

谁知道不定积分∫xln(x+1)dx是多少啊?∫xln(x-1)dx 利用分部积分法:=1/2∫ln(1+x)dx²=1/2x²ln(1+x)-1/2∫x&#

∫xln(x-1)dx∫xln(x-1)dx=1/2x²ln(1+x)-1/2[x²/2-x+ln(1+x)]+C。C为积分常数。解答

求不定积分∫xln(x+1)dx的解题步骤]dx=(1/2)x^2ln(x+1)-(1/2)[x^2/2-x+ln(x+1)]+C=(1/2)x^2ln(x+1)-(x^2

求不定积分∫xln(1+x)dx∫xln(1+x)dx的解答过程如下:

∫xIn(x-1)dx用分部积分法求解?+x-3/2+ln(x-1)+C 综上 得∫xIn(x-1)dx=1/2{x²In(x-1)-[1/2x²+x-3/2+ln(x-1)+C]}=(1/

求不定积分∫xln(x+1)dx∫xln(x+1)dx=∫ln(x+1)d(1/2*x^2)=1/2×x^2×ln(x+1)-1/2×∫x^2dln(

∫ln(x+1)dx分步积分u=x v=ln(x+1) u导=1 v导=1/(x+1)∫ln(x+1)dx=∫ln(x+1)d(x+1)=(ln(x+1))(x+1)-

用分布积分求∫xln(x-1)dxOK∫udv=uv-∫vdu知道吧这里:udv=xdx,v=(1/2)x^2所以:原式=[(1/2)x^2]ln(x-1

求∫xln(x+1)dx.解答:解:据题,有∫  xln(x+1)dx=x22ln(x+1) ∫x22 1x+1dx=x22ln(x+1

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