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不定积分练习题详解专题

不定积分练习题详解专题

Posted on 2026-02-17 17:21  K_J_M  阅读(0)  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 \(\frac{1}{3}[\intop \frac{1}{t}dt-\intop \frac{1}{t+1}d(t+1)]=\frac{1}{3}(\ln|x^3|-\ln|x^3+1|)+C\)