同样是分部积分，在∫xln(x+1)dx=(1/2)*∫ln(x+1)d(x^2+a)中, 取不同的 a 效果完全不一样，取 a=-1 效果最好。
提示:用分部积分法
看不清楚的话请点一下图片
原积分=(1/2)*∫ln(x+1)d(x^2) =(1/2)*[x^2*ln(x+1)-∫x^2*d(ln(x+1))] =(1/2)x^2*ln(x+1)-(1/2)∫x^2*[1/(x+1)]dx =(1/2)x^2*ln(x+1)-(1/2)∫[(x^2-1+1)/(x+1)]dx =(1/2)x^2*ln(x+1)-(1/2)∫(x-1)dx-(1/2)∫[1/(x+1)]dx =(1/2)x^2*ln(x+1)-(1/4)x^2+(1/2)x-(1/2)ln(x+1)+C =(1/2)*(x^2-1)*ln(x+1)-(1/4)x^2+(1/2)x+C.
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