摘要： 
\(\int x^{\alpha} \; \mathrm{d}x = \dfrac{x^{\alpha+1}}{\alpha+1}+C \space (\alpha \ne -1)\) \(\int \dfrac{1}{x} \; \mathrm{d}x = \ln x+C \space\) $\i 阅读全文
\(\int x^{\alpha} \; \mathrm{d}x = \dfrac{x^{\alpha+1}}{\alpha+1}+C \space (\alpha \ne -1)\) \(\int \dfrac{1}{x} \; \mathrm{d}x = \ln x+C \space\) $\i 阅读全文
posted @ 2024-10-31 13:27
Gokix
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