∫x除以x的3次方+1dx
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解决时间 2021-04-05 20:57
- 提问者网友：刺鸟
- 2021-04-05 01:19
∫x除以x的3次方+1dx
最佳答案
- 五星知识达人网友：行路难
- 2021-04-05 02:14
∫[x/(x³+1)]dx
=⅓∫[(x+1)/(x²-x+1) -1/(x+1)]dx
=(1/6)∫[(2x-1)/(x²-x+1) +3/(x²-x+1) - 2/(x+1)]dx
=(1/6)∫[(2x-1)/(x²-x+1)]dx+(√3/3)∫d(2x/√3 -1/√3)/[1+(2x/√3 - 1/√3)²] -⅓∫[1/(x+1)]dx
=(1/6)ln|x²-x+1|+(√3/3)arctan[√3(2x-1)/3] -⅓ln|x+1| +C
={ln[(x²-x+1)/(x²+2x+1)]+2√3arctan[√3(2x-1)/3]}/6 +C
=⅓∫[(x+1)/(x²-x+1) -1/(x+1)]dx
=(1/6)∫[(2x-1)/(x²-x+1) +3/(x²-x+1) - 2/(x+1)]dx
=(1/6)∫[(2x-1)/(x²-x+1)]dx+(√3/3)∫d(2x/√3 -1/√3)/[1+(2x/√3 - 1/√3)²] -⅓∫[1/(x+1)]dx
=(1/6)ln|x²-x+1|+(√3/3)arctan[√3(2x-1)/3] -⅓ln|x+1| +C
={ln[(x²-x+1)/(x²+2x+1)]+2√3arctan[√3(2x-1)/3]}/6 +C
全部回答
- 1楼网友：一袍清酒付
- 2021-04-05 05:18
=∫x/(x+1)(x²-x+1)dx
=-1/3∫1/(x+1)-(x+1)/(x²-x+1)dx
=(-1/3)ln|x+1|+1/6∫(2x-1+3)/(x²-x+1)dx
=(-1/3)ln|x+1|+(1/6)ln|x²-x+1|+1/2∫1/(x²-x+1)dx
然后∫1/(x²-x+1)dx=∫1/((x-1/2)²+3/4)dx
换元x=1/2+(√3/2)tanu，dx=(√3/2)sec²u
得积分=(2/√3)arctanu
=(2/√3)arctan((2x-1)/√3)
=-1/3∫1/(x+1)-(x+1)/(x²-x+1)dx
=(-1/3)ln|x+1|+1/6∫(2x-1+3)/(x²-x+1)dx
=(-1/3)ln|x+1|+(1/6)ln|x²-x+1|+1/2∫1/(x²-x+1)dx
然后∫1/(x²-x+1)dx=∫1/((x-1/2)²+3/4)dx
换元x=1/2+(√3/2)tanu，dx=(√3/2)sec²u
得积分=(2/√3)arctanu
=(2/√3)arctan((2x-1)/√3)
- 2楼网友：动情书生
- 2021-04-05 03:38
供参考。
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