第1个回答 2008-10-31
用分部积分公式
∫xln(x+1)dx=x^2ln(x+1)-∫[xln(x+1)+x^2/(x+1)]dx=x^2ln(x+1)-∫xln(x+1)dx-∫x^2/(x+1)dx……(1)
∫x^2/(x+1)dx=∫[(x+1)-2+1/(x+1)]dx=x^2/2-x+ln(x+1)+c
令∫xln(x+1)dx=y
由(1)式得y=x^2ln(x+1)-y-[x^2/2-x+ln(x+1)]
解出y=[x^2ln(x+1)]/2-[x^2/2-x+ln(x+1)]/2=-x^2/4+x/2-(x^2-1)[ln(x+1)]/2
第2个回答 2008-10-31
∫xln(x+1)dx
=∫ln(x+1)d(1/2*x^2)
=1/2×x^2×ln(x+1)-1/2×∫x^2dln(x+1)
=1/2×x^2×ln(x+1)-1/2×∫x^2/(x+1)dx
=1/2×x^2×ln(x+1)-1/2×∫[x-1+1/(x+1)]dx
=1/2×x^2×ln(x+1)-1/2×[1/2×x^2-x+ln(x+1)]+C
=1/2×(x^2-1)×ln(x+1)-1/4×(x^2-2x)+C本回答被提问者采纳
第3个回答 2019-06-18
【xlnx】′=1+lnx
所以对lnx积分=xlnx
-x
【x²lnx】=2xlnx+x所以对2xlnx积分=x²lnx-x²/2
∫xln(x-1)dx
=∫【(x-1)ln(x-1)+ln(x-1)】d(x-1)
分别积分
=0.5*(x-1)²ln(x-1)-0.25(x-1)²
+
(x-1)ln(x-1)-(x-1)+C
可以展开。思路就是这样。
或者xln(x-1)dx
=
1/2
ln(x-1)d(x²)
∫xln(x-1)dx
=1/2∫ln(x-1)d(x²)
=1/2【x²ln(x-1)-
∫x²*[1/(x-1)]dx】
1/2∫x²*[1/(x-1)]dx
=
1/2∫[x+1+1/(x-1)]dx
=
1/4x²+x/2+1/2ln(x-1)+
C
希望对你有帮助O(∩_∩)O~
强调一点,这里的x-1不能带绝对值,因为定义域就是x-1>0的。带绝对值扩大定义域了。
第4个回答 2019-01-08
分部积分法:∫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
其中
:∫1/2*x^2/(x-1)dx
分子-1,然后+1
,平方差公式,就容易积分,具体你自己去算算。