不定积分练习题详解专题

Posted on 2026-02-17 17:21  K_J_M  阅读(39)  评论(0)    收藏  举报

1.\(\intop \frac{1}{5x+3}dx=\frac{1}{5}\intop \frac{1}{5x+3}d(5x+3)=\frac{1}{5}\ln|5x+3|+C\)

2.\(\intop xe^{x^2}dx=\frac{1}{2}\intop e^{x^2}d(x^2)=\frac{1}{2}e^{x^2}+C\)

3.\(\intop x\sqrt{1-x^2}dx=\frac{1}{2}\intop \sqrt{1-x^2}d(x^2)\),令 \(x=\sin t\)\(t \in (-\frac{\pi}{2},\frac{\pi}{2})\),则有 \(\frac{1}{2}\intop \cos t\times 2\sin t \cos tdt=\intop \sin t\cos^2 tdt=-\intop \cos^2 td(\cos t)=\frac{1}{3}\cos^3 (\arcsin x)+C\)

4.\(\intop \frac{1}{x^2}\sin \frac{1}{x}dx=\intop -\sin \frac{1}{x}d(\frac{1}{x})=\cos \frac{1}{x}+C\)

5.\(\intop \frac{e^{3\sqrt{x}}}{\sqrt{x}}dx=2\intop e^{3\sqrt{x}}d(\sqrt{x})=2\intop e^{2\sqrt{x}}d(e^{\sqrt{x}})=\frac{2}{3}e^{3\sqrt{x}}+C\)

6.\(\intop \frac{1}{x(1+x^6)}dx=\intop \frac{x^2}{x^3(1+x^6)}dx=\frac{1}{3}\intop \frac{d(x^3)}{x^3(1+x^6)}\),令 \(x^3=t\),则 \(\frac{1}{3}\intop \frac{dt}{t(1+t^2)}=\frac{1}{3}\intop \frac{1}{t}dt-\frac{1}{3}\intop \frac{t}{1+t^2}dt=\frac{1}{3}\ln|x^3|-\frac{1}{6}\ln(x^2+1)+C\)

7.\(\intop \cos 2xdx=\frac{1}{2}\intop \cos2xd(2x)=\frac{1}{2}\sin 2x+C\)

8.\(\intop \frac{\sin x}{\sqrt{5+\cos x}}dx=-\intop \frac{1}{\sqrt{5+\cos x}}d(\cos x)\),令 \(\cos x =t\),则 \(-\intop \frac{1}{\sqrt{5+t}}d(5+t)=-2\intop d(\sqrt{5+t})=-2\sqrt{5+\cos x}+C\)

9.\(\intop \tan^4 xdx=\intop (\sec^2 x-1)^2dx=\intop \sec^4dx-2\intop \sec^2 xdx+\intop dx=\intop (\tan^2 x+1) d(\tan x)-2\tan x+x=\frac{1}{3}\tan^3 x+\tan x-2\tan x+x=\frac{1}{3}\tan^3 x-\tan x+x+C\)

10.令 \(t=e^x\),则 \(x=\ln t\)\(\intop \frac{e^{2x}}{1+e^x}dx=\intop \frac{t^2}{1+t}d\ln t=\intop \frac{t}{1+t}dt=e^x-\ln|1+e^x|+C\)

11.\(\intop \frac{1}{1+e^x}dx=x-\ln|e^x+1|+C\)

12.\(\intop \frac{1}{x\ln^2 x}dx=\intop \frac{1}{\ln^2 x}d(\ln x)=-\frac{1}{\ln x}+C\)

13.\(\intop \frac{1}{x(1+2\ln x)}dx=\frac{1}{2}\intop \frac{1}{1+2\ln x}d(2\ln x+1)=\frac{1}{2}\ln|2\ln x+1|+C\)