求定积分上限为1下限为0x㏑﹙x+1﹚dx=求定积分上限为1下限为0㏑﹙x+1﹚dx^2/2

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/30 19:06:43
求定积分上限为1下限为0x㏑﹙x+1﹚dx=求定积分上限为1下限为0㏑﹙x+1﹚dx^2/2
x){YϗdG˙edG7eP3+ RaKH=8#}#"}zX_`gC7

求定积分上限为1下限为0x㏑﹙x+1﹚dx=求定积分上限为1下限为0㏑﹙x+1﹚dx^2/2
求定积分上限为1下限为0x㏑﹙x+1﹚dx=求定积分上限为1下限为0㏑﹙x+1﹚dx^2/2

求定积分上限为1下限为0x㏑﹙x+1﹚dx=求定积分上限为1下限为0㏑﹙x+1﹚dx^2/2
∫x㏑﹙x+1﹚dx
=1/2∫ln(x+1)d(x^2)
=1/2[x^2ln(x+1)-∫x^2/(x+1)dx]
=1/2[x^2ln(x+1)-(∫(x-1+1/(x+1))dx)]
=1/2[x^2ln(x+1)-(x^2/2 -x +ln(x+1))]
=1/2[(ln2)-(1/2-1+ln2) ]
=1/4