解析 ∫ln x\,dx = xln x - x + C 本题使用分部积分法求解。令 u = ln x,dv = dx,则 du = 1/xdx,v = x。根据分部积分公式,原式可化为: ∫ln x\,dx = xln x - ∫ x⋅1/xdx = xln x - ∫ dx = xln x - x + C。 因此,∫ln x\,dx = xln x - x + C。
令$u = \ln x$,则 $du = \frac{1}{x} dx$, 将这两个量带入积分式中,得到: $\int \frac{1}{u} du$。 对$\int \frac{1}{u} du$ 进行积分计算: $\int \frac{1}{u} du = \ln |u| + C = \ln |\ln x| + C$。 所以不定积分 $\int \frac{1}{x\ln x} dx = \...
Integrate: {eq}\int x^n \ln x dx{/eq} Method of Integration by Parts: Integration by parts is a special method of integration that is very often useful when two integrable functions are multiplied together. If we suppose that {eq}f {/eq} and {eq}g {/eq} are two real valued fun...
Answer to: Integrate by substitution: \int (e^{ln x) }dx/x. By signing up, you'll get thousands of step-by-step solutions to your homework...
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利用分部积分法,\( \int x \ln x dx \) 可以分解为 \( \ln x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} dx \),其结果为 ___。相关知识点: 试题来源: 解析 答案:\( \ln x \cdot \frac{x^2}{2} - \frac{x^3}{6} + C \) 反馈 收藏...
百度试题 结果1 题目计算:∫_1^1xlnxdx\int _{0}^{1}x \ln xdx 相关知识点: 试题来源: 解析 原式-∫_0^1xlnxdx=((x^2)/2lnx-(x^2)/4)^1=(0-1/4)-(0-0)=-1/4;综上所述,答案为:-1/4 反馈 收藏
∫x(logx)2dx View Solution ∫[logx−11+(logx)2]2dx = View Solution Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma Solutions for Physics ...
Integrate: {eq}\int \frac{dx}{x \ln x} {/eq} Integration By Substitution: If we have an integral in the form {eq}\int f(g(x))g'(x) dx {/eq}, then integration by substitution is the best method to solve this type of integral. ...
Answer to: Find the integral. \int x ln x dx. By signing up, you'll get thousands of step-by-step solutions to your homework questions. You can...