ln²(x+1) +x *2ln(x+1) *[ln(x+1)]'=ln²(x+1) +2xln(x+1) /(x+1)而如果是x^ln(x+1)就先用对数恒等式，得到e^[lnx *ln(x+1)]再求导
∫xln(1+x)dx 相关知识点: 试题来源: 解析 反对幂指三 前面设为u,后面的设为v' (最后两个随意)令u=ln(x+1) ,v'=x∫xln(1+x) dx=(x2/2)ln(1+x)-∫(x2/2)/(1+x) dx=(x2/2)ln(1+x)-∫[(x2/2)-(1/2)+(1/2)]/(1+x) dx=(x2/2)ln(1+x)-∫[(x-1)/2+1/2(1...
百度试题 题目思路: 分部积分。 ★★(2) xln(x 1相关知识点: 试题来源: 解析 错误 反馈 收藏
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∫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×(...
∫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... 解析看不懂?免费查看同类题视频...
原式=1/2∫ln(1+x)dx²=1/2x²ln(1+x)-1/2∫x²dln(1+x)=1/2x²ln(1+x)-1/2∫x²/(1+x) dx=1/2x²ln(1+x)-1/2∫(x²-1+1)/(1+x) dx=1/2x²ln(1+x)-1/2∫[(x²-1)/(x+1)+1/(1+x)] dx=1/2x²ln(1+x)-1/2∫[(x-1)+1/(1+x)] dx...
百度试题 结果1 题目【题目】f(x)=xln(1+x)求的导数 相关知识点: 试题来源: 解析 【解析】这理: ln(1+x)^2=((1+x)^1)/(1+x)=1/(1+x) 反馈 收藏
原式=1/2∫ln(1+x)dx²=1/2x²ln(1+x)-1/2∫x²dln(1+x)=1/2x²ln(1+x)-1/2∫x²/(1+x) dx=1/2x²ln(1+x)-1/2∫(x²-1+1)/(1+x) dx=1/2x²ln(1+x)-1/2∫[(x²-1)/(x+1)+1/(1+x)] dx=1/2x²ln(1+x)-1/2∫[(x-1)+1/(1+x)] dx...
简单分析一下，答案如图所示 ∫