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
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...
xln(1+x)的不定积分结果为:(1/2)x²ln(1+x) - (1/2)(x² - 2x + 2ln(1+x)) + C。以下是具体
简单计算一下即可，答案如图所示
简单分析一下，答案如图所示
∫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... 结果...
答案: 1/2(x^2-1)ln(1+x)-1/4x^2+1/2x+c 解析: ∫xln(x+1)dx =∫ln(1+x)d(1/2x^2) =1/2x^2ln(t+x)-1/2∫(x^2)/(1+x)dx =1/2x^2ln(t+x)-1/2∫(x-1)dx-1/2∫1/(1+x)dx =1/2(x^2-1)ln(1+x)-1/4x^2+1/2x+c 知识点:本题考查不定积分的分部积分法 ...
∫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...
应用分部积分法公式∫udv=uv-∫vdu，得：∫xln(x+1)dx=x[1/2(x+1)ln(x+1)-1/2(x+1)]-∫[1/2(x+1)ln(x+1)-1/2(x+1)]dx 进一步简化：=x[1/2(x+1)ln(x+1)-1/2(x+1)]-1/2∫(x+1)ln(x+1)dx+1/2∫(x+1)dx 继续化简：=1/2x(x+1)ln(x+1)-1/2x(x...
∫xln（1+x）dx的解答过程如下：