2025年凑微分考试题目及答案

2025年凑微分考试题目及答案

一、单项选择题1.下列凑微分正确的是（）A.\(xdx=\frac{1}{2}d(x^2)\)B.\(e^{2x}dx=d(e^{2x})\)C.\(\frac{1}{x}dx=d(\frac{1}{x^2})\)D.\(\sinxdx=d(\cosx)\)答案：A2.若\(\intf(x)dx=F(x)+C\)，则\(\intf(3x+2)dx\)等于（）A.\(F(3x+2)+C\)B.\(\frac{1}{3}F(3x+2)+C\)C.\(3F(3x+2)+C\)D.\(F(x)+C\)答案：B3.凑微分\(\int\frac{1}{\sqrt{1-4x^2}}dx\)，正确的是（）A.\(\frac{1}{2}\int\frac{1}{\sqrt{1-(2x)^2}}d(2x)\)B.\(\int\frac{1}{\sqrt{1-(2x)^2}}d(2x)\)C.\(2\int\frac{1}{\sqrt{1-(2x)^2}}d(2x)\)D.\(\frac{1}{4}\int\frac{1}{\sqrt{1-(2x)^2}}d(2x)\)答案：A4.对于积分\(\intxe^{x^2}dx\)，凑微分后为（）A.\(\frac{1}{2}\inte^{x^2}d(x^2)\)B.\(\inte^{x^2}d(x^2)\)C.\(2\inte^{x^2}d(x^2)\)D.\(\frac{1}{2}\inte^{x^2}dx\)答案：A5.若要将\(\int\cos(5x-3)dx\)凑微分，结果是（）A.\(\frac{1}{5}\int\cos(5x-3)d(5x-3)\)B.\(\int\cos(5x-3)d(5x-3)\)C.\(5\int\cos(5x-3)d(5x-3)\)D.\(\frac{1}{3}\int\cos(5x-3)d(5x-3)\)答案：A6.凑微分\(\int\frac{1}{x^2}e^{\frac{1}{x}}dx\)，得到（）A.\(-\inte^{\frac{1}{x}}d(\frac{1}{x})\)B.\(\inte^{\frac{1}{x}}d(\frac{1}{x})\)C.\(-\inte^{\frac{1}{x}}dx\)D.\(\inte^{\frac{1}{x}}dx\)答案：A7.已知\(\intf(x)dx=\sinx+C\)，则\(\intf(2x)dx\)为（）A.\(\sin(2x)+C\)B.\(\frac{1}{2}\sin(2x)+C\)C.\(2\sin(2x)+C\)D.\(\sinx+C\)答案：B8.对于积分\(\int\frac{1}{1+9x^2}dx\)，凑微分正确的是（）A.\(\frac{1}{3}\int\frac{1}{1+(3x)^2}d(3x)\)B.\(\int\frac{1}{1+(3x)^2}d(3x)\)C.\(3\int\frac{1}{1+(3x)^2}d(3x)\)D.\(\frac{1}{9}\int\frac{1}{1+(3x)^2}d(3x)\)答案：A9.凑微分\(\int\sec^2(2x+1)dx\)，结果是（）A.\(\frac{1}{2}\int\sec^2(2x+1)d(2x+1)\)B.\(\int\sec^2(2x+1)d(2x+1)\)C.\(2\int\sec^2(2x+1)d(2x+1)\)D.\(\frac{1}{4}\int\sec^2(2x+1)d(2x+1)\)答案：A10.若\(\intf(x)dx=\lnx+C\)，则\(\intf(4x)dx\)等于（）A.\(\ln(4x)+C\)B.\(\frac{1}{4}\ln(4x)+C\)C.\(4\ln(4x)+C\)D.\(\lnx+C\)答案：B二、多项选择题1.下列凑微分正确的有（）A.\(2xdx=d(x^2)\)B.\(\cosxdx=d(\sinx)\)C.\(e^xdx=d(e^x)\)D.\(\frac{1}{x^2}dx=d(\frac{1}{x})\)答案：ABC2.对于积分\(\intg(ax+b)dx\)（\(a\neq0\)），可凑微分为（）A.\(\frac{1}{a}\intg(ax+b)d(ax+b)\)B.\(\intg(ax+b)d(ax+b)\)C.\(a\intg(ax+b)d(ax+b)\)D.令\(u=ax+b\)，\(du=adx\)，则\(\intg(ax+b)dx=\frac{1}{a}\intg(u)du\)答案：AD3.下列能通过凑微分求解的积分有（）A.\(\intx^3e^{x^4}dx\)B.\(\int\frac{\sin\sqrt{x}}{\sqrt{x}}dx\)C.\(\int\frac{1}{x\lnx}dx\)D.\(\int\cos^2xdx\)答案：ABC4.凑微分\(\int\frac{1}{\sqrt{x}(1+x)}dx\)，可采用的方法有（）A.令\(t=\sqrt{x}\)，\(dt=\frac{1}{2\sqrt{x}}dx\)，则\(2\int\frac{1}{1+t^2}dt\)B.把\(\frac{1}{\sqrt{x}(1+x)}\)变形为\(2\frac{1}{2\sqrt{x}(1+(\sqrt{x})^2)}\)，凑出\(d(\sqrt{x})\)C.\(\int\frac{1}{\sqrt{x}(1+(\sqrt{x})^2)}d(\sqrt{x})\)D.\(\int\frac{1}{1+x}d(\sqrt{x})\)答案：ABC5.已知\(\intf(x)dx=F(x)+C\)，则（）A.\(\intf(ax+b)dx=\frac{1}{a}F(ax+b)+C\)（\(a\neq0\)）B.\(\intf(x^n)nx^{n-1}dx=F(x^n)+C\)C.\(\intf(e^x)e^xdx=F(e^x)+C\)D.\(\intf(\sinx)\cosxdx=F(\sinx)+C\)答案：ABCD6.下列凑微分错误的有（）A.\(\int\frac{1}{x^3}dx=d(-\frac{1}{2x^2})\)B.\(\int\tanxdx=d(\ln|\cosx|)\)C.\(\int\frac{1}{\sqrt{4-x^2}}dx=d(\arcsin\frac{x}{2})\)D.\(\int\frac{1}{1+x^2}dx=d(\arctan\frac{1}{x})\)答案：BD7.对于积分\(\int\sin(ax+b)\cos(ax+b)dx\)（\(a\neq0\)），可以采用的凑微分方法有（）A.令\(u=\sin(ax+b)\)，\(du=a\cos(ax+b)dx\)，\(\frac{1}{a}\intudu\)B.令\(v=\cos(ax+b)\)，\(dv=-a\sin(ax+b)dx\)，\(-\frac{1}{a}\intvdv\)C.利用\(\sin2\alpha=2\sin\alpha\cos\alpha\)，将积分变为\(\frac{1}{2a}\int\sin(2(ax+b))d(2(ax+b))\)D.直接凑出\(d(\sin(ax+b)\cos(ax+b))\)答案：ABC8.下列积分中，通过凑微分能转化为基本积分公式的有（）A.\(\int\frac{e^{2x}}{1+e^{2x}}dx\)B.\(\int\frac{\cosx}{1+\sin^2x}dx\)C.\(\int\frac{1}{x^2+2x+2}dx\)D.\(\int\frac{1}{\sqrt{4x-x^2}}dx\)答案：ABC9.凑微分\(\int\frac{x^2}{(x^3+1)^2}dx\)，正确的是（）A.\(\frac{1}{3}\int\frac{1}{(x^3+1)^2}d(x^3+1)\)B.令\(u=x^3+1\)，\(du=3x^2dx\)，\(\frac{1}{3}\int\frac{1}{u^2}du\)C.\(\int\frac{1}{(x^3+1)^2}d(x^3)\)D.\(\frac{1}{3}\int\frac{1}{(x^3+1)^2}dx^3\)答案：ABD10.已知\(\intf(x)dx=F(x)+C\)，对于\(\intf(ax^2+bx+c)(2ax+b)dx\)（\(a\neq0\)），正确的是（）A.令\(u=ax^2+bx+c\)，\(du=(2ax+b)dx\)，则积分结果为\(F(ax^2+bx+c)+C\)B.可以凑微分为\(\intf(ax^2+bx+c)d(ax^2+bx+c)\)C.与\(F(x)\)没有关系D.是\(F(ax^2+bx+c)\)的原函数答案：ABD三、判断题1.\(3x^2dx=d(x^3)\)（）答案：对2.\(\intf(2x)dx=2\intf(2x)d(2x)\)（）答案：错3.凑微分\(\int\frac{1}{x^2+1}dx=d(\arctanx)\)（）答案：错4.若\(\intf(x)dx=F(x)+C\)，则\(\intf(3-2x)dx=-\frac{1}{2}F(3-2x)+C\)（）答案：对5.\(\int\sin2xdx=\frac{1}{2}\int\sin2xd(2x)\)（）答案：对6.凑微分\(\int\frac{1}{\sqrt{1-x^2}}dx=d(\arccosx)\)（）答案：错7.已知\(\intf(x)dx=F(x)+C\)，那么\(\intf(x+1)dx=F(x+1)+C\)（）答案：对8.\(\int\frac{1}{x\sqrt{x^2-1}}dx=d(\arccos\frac{1}{x})\)（）答案：错9.对于积分\(\int\cos^3x\sinxdx\)，令\(u=\cosx\)，\(du=-\sinxdx\)，则积分可化为\(-\intu^3du\)（）答案：对10.\(\int\frac{1}{e^x+1}dx\)不能通过凑微分求解（）答案：错四、简答题1.简述凑微分法的基本原理。凑微分法基于复合函数求导的逆运算。若\(F^\prime(u)=f(u)\)，且\(u=\varphi(x)\)可导，那么\([F(\varphi(x))]^\prime=f(\varphi(x))\varphi^\prime(x)\)。凑微分就是把被积函数中一部分凑成某个函数的微分形式，使得积分能转化为已知的积分公式形式来求解，关键在于找到合适的\(u\)以及\(du\)与\(dx\)的关系。2.说明如何将\(\int\frac{1}{x^2+4x+5}dx\)进行凑微分求解。先对分母进行变形，\(x^2+4x+5=x^2+4x+4+1=(x+2)^2+1\)。令\(u=x+2\)，则\(du=dx\)。原积分\(\int\frac{1}{x^2+4x+5}dx\)就变为\(\int\frac{1}{u^2+1}du\)，根据基本积分公式\(\int\frac{1}{u^2+1}du=\arctanu+C\)，再把\(u=x+2\)代回，结果为\(\arctan(x+2)+C\)。3.对于积分\(\int\sin3x\cos2xdx\)，怎样利用凑微分法求解？利用三角函数积化和差公式\(\sinA\cosB=\frac{1}{2}[\sin(A+B)+\si