x^4/(x+1)的不定积分
答案:4 悬赏:20 手机版
解决时间 2021-02-15 19:25
- 提问者网友:那叫心脏的地方装的都是你
- 2021-02-15 12:18
x^4/(x+1)的不定积分
最佳答案
- 五星知识达人网友:蕴藏春秋
- 2021-02-15 13:31
x⁴/(x+1)=(x⁴-1+1)/(x+1)
=[(x²-1)(x²+1)]/(x+1) +1/(x+1)
=(x-1)(x+1)(x²+1)/(x+1) +1/(x+1)
=(x-1)(x²+1) +1/(x+1)
=x³-x²+x-1 +1/(x+1)
∫x⁴/(x+1) dx
=∫[x³-x²+x-1 +1/(x+1)]dx
=x⁴/4 -x³/3 +x/2 -x +ln(x+1)+C
全部回答
- 1楼网友:行路难
- 2021-02-15 15:52
解:
x^4/(x+1)=x³-x²+x+1/(x+1)-1
∫x^4/(x+1) dx
=∫[x³-x²+x+1/(x+1)-1]dx
=∫x³dx-∫x²dx+∫xdx+∫1/(x+1)-∫1dx
=x^4/4-x³/3+x²/2+ln|x+1|-x+C
- 2楼网友:人间朝暮
- 2021-02-15 15:20
解: ∫arctan(x+1)dx =xarctan(x+1)-∫xdarctan(x+1) =xarctan(x+1)-∫x*1/[1+(x+1)^2]dx =xarctan(x+1)-∫x/(x^2+2x+2)dx =xarctan(x+1)-1/2∫(2x+2)/(x^2+2x+2)dx+∫1/(x^2+2x+2)dx =xarctan(x+1)-1/2∫(2x+2)/(x^2+2x+2)dx+∫1/[1+(x+1)^2]d(x+1) =x...
- 3楼网友:孤独入客枕
- 2021-02-15 14:38
x^4/(x+1)
=[(x+1)-1]^4/(x+1)
=[(x+1)^4-4(x+1)^3+6(x+1)^2-4(x+1)+1]/(x+1)
=(x+1)^3-4(x+1)^2+6(x+1)-4+1/(x+1)
原式=∫[(x+1)^3-4(x+1)^2+6(x+1)-4+1/(x+1)]dx
=1/4*(x+1)^4-4/3*(x+1)^3+3(x+1)^2-4(x+1)+ln|x+1|+C
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