∫﹛﹙x-1﹚/﹙x2+x+1﹚﹜dx怎么做
答案:1 悬赏:30 手机版
解决时间 2021-03-11 17:00
- 提问者网友:沉默菋噵
- 2021-03-10 18:57
∫﹛﹙x-1﹚/﹙x2+x+1﹚﹜dx怎么做
最佳答案
- 五星知识达人网友:酒醒三更
- 2021-03-10 19:47
∫ (x - 1)/(x² + x + 1) dx
= ∫ {(1/2)[(2x + 1) - 1] - 1}/(x² + x + 1) dx
= (1/2)∫ (2x + 1)/(x² + x + 1) dx - (3/2)∫ dx/(x² + x + 1)
= (1/2)∫ d(x² + x + 1)/(x² + x + 1) - (3/2)∫ d(x + 1/2)/[(x + 1/2)² + 3/4]
= (1/2)ln|x² + x + 1| - (3/2)(2/√3)arctan[(x + 1/2)(2/√3)] + C
= (1/2)ln|x² + x + 1| - √3arctan[(2x + 1)/√3] + C
= ∫ {(1/2)[(2x + 1) - 1] - 1}/(x² + x + 1) dx
= (1/2)∫ (2x + 1)/(x² + x + 1) dx - (3/2)∫ dx/(x² + x + 1)
= (1/2)∫ d(x² + x + 1)/(x² + x + 1) - (3/2)∫ d(x + 1/2)/[(x + 1/2)² + 3/4]
= (1/2)ln|x² + x + 1| - (3/2)(2/√3)arctan[(x + 1/2)(2/√3)] + C
= (1/2)ln|x² + x + 1| - √3arctan[(2x + 1)/√3] + C
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