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高数不定积分72题解答

news 2025/6/23 17:12:47

题目来源：这72道积分题目会积了，绝对是高高手
作者：湖心亭看雪

第一题

$原式=∫15x+3dx=15∫15x+3d(5x+3)=15ln(5x+3)+C\begin{aligned} \text{原式}&=\int \frac{1}{5x+3}dx \\ &=\frac{1}{5} \int\frac{1}{5x+3}d(5x+3) \\ &=\frac{1}{5} ln(5x+3) +C \end{aligned}$

第二题

$原式=∫e2x+3dx=12∫e2x+3d(2x+3)=12e2x+3+C\begin{aligned} \text{原式}&=\int e^{2x+3}dx \\ &=\frac{1}{2} \int e^{2x+3}d(2x+3) \\ &=\frac{1}{2} e^{2x+3}+C \end{aligned}$

第三题

$原式=∫xex2dx=12∫ex2dx2=12ex2+C\begin{aligned} \text{原式}&=\int xe^{x^2} dx \\ &= \frac{1}{2} \int e^{x^2} dx^2 \\ &=\frac{1}{2}e^{x^2}+C \end{aligned}$

第四题

$原式=∫x1−x2dx=−12∫1−x2d(1−x2)=−13(1−x2)32+C\begin{aligned} \text{原式}&=\int x\sqrt{1-x^2} dx \\ &=-\frac{1}{2} \int \sqrt{1-x^2}d(1-x^2) \\ &=-\frac{1}{3}(1-x^2)^{\frac{3}{2} } +C \end{aligned}$

第五题

$原式=∫1x2sin1xdx=−∫sin1xd(1x)=cos(1x)+C\begin{aligned} \text{原式}&=\int \frac{1}{x^2}sin{\frac{1}{x}} dx \\ &=-\int sin{\frac{1}{x}}d(\frac{1}{x} ) \\ &=cos(\frac{1}{x})+C \end{aligned}$

第六题

$原式=∫e3xxdx=2∫e3x2xdx=23∫e3xd(3x)=23e3x+C\begin{aligned} \text{原式}&=\int \frac{e^{3\sqrt[]{x} }}{\sqrt[]{x}} dx \\ &=2\int \frac{e^{3\sqrt[]{x} }}{2\sqrt{x} } dx \\ &=\frac{2}{3} \int e^{3\sqrt[]{x} }d(3\sqrt[]{x}) \\ &=\frac{2}{3}e^{3\sqrt[]{x}}+C \end{aligned}$

第七题

$原式=∫1x(1+x6)dx=∫(1x−x51+x6)dx=lnx−16∫11+x6d(1+x6)=lnx−16ln(1+x6)+C\begin{aligned} \text{原式}&=\int \frac{1}{x(1+x^6)} dx \\ &=\int (\frac{1}{x}-\frac{x^5}{1+x^6} )dx \\ &=lnx-\frac{1}{6} \int \frac{1}{1+x^6} d(1+x^6) \\ &=lnx-\frac{1}{6}ln(1+x^6)+C \end{aligned}$

第八题

$原式=∫cos2xdx=12∫cos2xd2x=12sin2x+C\begin{aligned} \text{原式}&=\int cos2x dx \\ &=\frac{1}{2} \int cos2x d2x \\ &=\frac{1}{2}sin2x +C \end{aligned}$

第九题

$原式=∫sinx5+cosxdx=−∫15+cosxd(5+cosx)=−2∫125+cosxd(5+cosx)=−25+cosx+C\begin{aligned} \text{原式}&=\int \frac{sinx}{\sqrt[]{5+cosx} }dx \\ &=-\int \frac{1}{\sqrt[]{5+cosx}} d(5+cosx) \\ &=-2\int \frac{1}{2~\sqrt[]{5+cosx}}d(5+cosx) \\ &=-2~\sqrt[]{5+cosx}+C \end{aligned}$

第十题

$原式=∫tan4xdx=∫(sec2x−1)2dx=∫(sec4x−2sec2x+1)dx=∫sec2x(sec2x−2)dx+∫1dx=∫sec2x(tan2x−1)dx+x=∫sec2xtan2xdx−∫sec2dx+x=∫tan2xdtanx−tanx+x=13tan3x−tanx+x+C\begin{aligned} \text{原式}&=\int tan^4xdx \\ &=\int(sec^2x-1)^2dx \\ &=\int (sec^4x-2sec^2x+1)dx \\ &=\int sec^2x(sec^2x-2) dx+\int 1dx \\ &=\int sec^2x(tan^2x-1) dx+x \\ &=\int sec^2xtan^2xdx-\int sec^2 dx+x \\ &=\int tan^2x dtanx -tanx+x \\ &=\frac{1}{3}tan^3x -tanx +x+C \end{aligned}$

第十一题

$原式=∫e2x1+exdx=∫e2x−1+11+exdx=∫(ex−1)(ex+1)+11+exdx=∫(ex−1)dx+∫11+exdx=ex−x+∫1+ex−ex1+exdx=ex−x+∫1dx−∫ex1+exdx=ex−∫11+exdex=ex−ln(1+ex)+C\begin{aligned} \text{原式}&=\int \frac{e^{2x}}{1+e^x} dx \\ &=\int\frac{e^{2x}-1+1}{1+e^x} dx \\ &=\int\frac{(e^x-1)(e^x+1)+1}{1+e^x} dx \\ &=\int (e^x-1)dx+\int \frac{1}{1+e^x} dx \\ &=e^x-x+\int \frac{1+e^x-e^x}{1+e^x} dx \\ &=e^x-x+\int 1 dx-\int \frac{e^x}{1+e^x} dx \\ &=e^x-\int \frac{1}{1+e^x} de^x \\ &=e^x-ln(1+e^x)+C \end{aligned}$

第十二题

$原式=∫11+exdx=∫1+ex−ex1+exdx=∫1dx−∫ex1+exdx=x−∫11+exdex=x−ln(1+ex)+C\begin{aligned} \text{原式}&=\int \frac{1}{1+e^x} dx \\ &=\int \frac{1+e^x-e^x}{1+e^x} dx \\ &=\int 1 dx-\int \frac{e^x}{1+e^x} dx \\ &=x-\int \frac{1}{1+e^x} de^x \\ &=x-ln(1+e^x)+C \end{aligned}$

第十三题

$原式=∫1xln2x=∫1ln2xdlnx=−1lnx+C\begin{aligned} \text{原式}&=\int \frac{1}{xln^2x} \\ &= \int \frac{1}{ln^2x}dlnx \\ &= -\frac{1}{lnx} + C \end{aligned}$

第十四题

$原式=∫1x(1+2lnx)dx=12∫11+2lnxd(2lnx)=12ln(1+2lnx)+C\begin{aligned} \text{原式}&=\int \frac{1}{x(1+2lnx)}dx \\ &=\frac{1}{2} \int \frac{1}{1+2lnx} d(2lnx) \\ &=\frac{1}{2}ln(1+2lnx)+C \end{aligned}$

第十五题

$原式=∫1a2cos2x+b2sin2xdx=∫1cos2xa2+b2tan2xdx=∫sec2xa2+b2tan2xdx=∫1a2+b2tan2xdtanx=1ab∫11+b2tan2xa2d(btanxa)=1abarctan(btanxa)+C\begin{aligned} \text{原式}&=\int \frac{1}{a^2cos^2x+b^2sin^2x}dx \\ &=\int \frac{\frac{1}{cos^2x} }{a^2+b^2tan^2x}dx \\ &=\int \frac{sec^2x }{a^2+b^2tan^2x}dx \\ &=\int \frac{1}{a^2+b^2tan^2x}dtanx \\ &=\frac{1}{ab} \int \frac{1}{1+\frac{b^2tan^2x}{a^2} }d(\frac{btanx}{a}) \\ &=\frac{1}{ab}arctan(\frac{btanx}{a}) +C \end{aligned}$