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楼主 |
发表于 2021-1-7 03:13
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题:计算\(\,\displaystyle\int_{1/\alpha}^\alpha\frac{\arctan x}{x}dx\;(\alpha>1)\)
解:\(\,\displaystyle\int_{1/\alpha}^\alpha\frac{\arctan x}{x}dx\overset{t=1/x}{=}\int_{1/\alpha}^{\alpha}\frac{\arctan\frac{1}{t}}{t}dt,\;\;\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}\)
\(\therefore\;\;\displaystyle\int_{1/\alpha}^{\alpha}\frac{\arctan x}{x}dx=\frac{\pi}{4}\int_{1/\alpha}^{\alpha}\frac{dx}{x}=\frac{\pi}{2}\ln\alpha.\) |
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