求不定积分∫xln(x+1)dx

∫xln(x+1)dx
=∫(x+1)ln(x+1)d(x+1)-∫ln(x+1)d(x+1)
=0.5(∫ln(x+1)d(x+1)^2-∫ln(x+1)d(x+1))
=0.5((x+1)^2ln(x+1)-∫(x+1)^2dln(x+1)-(x+1)ln(x+1)+∫(x+1)dln(x+1))
=0.5((x+1)^2ln(x+1)-∫(x+1)dx-(x+1)ln(x+1)+∫dx)
=0.5((x+1)^2ln(x+1)-0.5x^2-x-(x+1)ln(x+1)+x)
=0.5(x+1)^2ln(x+1)-0.25x^2-0.5(x+1)ln(x+1)+C

这题要采用分部积分法
xln(x-1)dx
=
ln(x-1)d(x²)
∫xln(x－1)dx
=∫ln(x－1)d(x²)
=x²ln(x-1)-
∫x²*[1/(x-1)]dx
∫x²*[1/(x-1)]dx
=
∫[x+1+1/(x-1)]dx
=
1/2x²+x+ln|x-1|
+
C
仅供参考~

xln(x－1）dx
=1/2∫ln(x－1）dx^2
=1/2x^2*ln(x－1）-∫1/2*x^2/(x-1)dx
=1/2x^2*ln(x－1）-1/4
*x^2-1/2x
-ln(x－1）+C

解答如下图片：

用分布积分公式
∫uv'=uv-∫u'v 把x看成u ln(x+1)看成v
所以原式=(x*x/2)*ln(x+1)-(1/2)∫(x*x)/(x+1)dx
再看∫(x*x)/(x+1)dx=∫[(x+1)(x-1)+1]/(x+1)dx
=∫[(x-1)+1/(x+1)]dx
=∫(x-1)dx+∫1/(x+1)dx
=∫xdx-∫dx+∫1/(x+1)d(x+1)
=1/(2x*x)-x+ln|x+1|
把这个结果代入上式即可