求有理函数不定积分∫(x-1)/(x²+2x+3)dx有没简单点的办法
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![求有理函数不定积分∫(x-1)/(x²+2x+3)dx有没简单点的办法](/uploads/image/z/3836922-42-2.jpg?t=%E6%B1%82%E6%9C%89%E7%90%86%E5%87%BD%E6%95%B0%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%28x-1%29%2F%28x%26%23178%3B%2B2x%2B3%29dx%E6%9C%89%E6%B2%A1%E7%AE%80%E5%8D%95%E7%82%B9%E7%9A%84%E5%8A%9E%E6%B3%95)
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求有理函数不定积分∫(x-1)/(x²+2x+3)dx有没简单点的办法
求有理函数不定积分
∫(x-1)/(x²+2x+3)dx有没简单点的办法
求有理函数不定积分∫(x-1)/(x²+2x+3)dx有没简单点的办法
改一下如下图
∫(x-1)/(x²+2x+3)dx
=(1/2)∫(2x+2)/(x²+2x+3)dx-2∫1/(x²+2x+1+2)dx
=(1/2)ln(x²+2x+3)-(√2)arctan[(x+1)/√2]+C