∫dx/[x(1+x^8)]=_____
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解决时间 2021-03-15 18:30
- 提问者网友：我们很暧昧
- 2021-03-14 18:01
∫dx/[x(1+x^8)]=_____
最佳答案
- 五星知识达人网友：大漠
- 2021-03-14 19:21
1-x^8 = -(1+x^8) +2
∫(1-x^8)/[x(1+x^8)] dx
=-∫(1/x) dx + 2∫dx/[x(1+x^8)]
=-ln|x| +2∫dx/[x(1+x^8)]
let
x^4 = tany
4x^3 dx = (secy)^2 .dy
∫dx/[x(1+x^8)]
=(1/4)∫(4x^3 dx)/[x^4.(1+x^8)]
=(1/4)∫ (secy)^2 .dy/[tany(secy)^2]
=(1/4)∫ dy/tany
=(1/4)ln|siny| + C'
=(1/4)ln| x^4/√(1+x^8)| + C'
∫(1-x^8)/[x(1+x^8)] dx
=-ln|x| +2∫dx/[x(1+x^8)]
=-ln|x| +(1/2)ln| x^4/√(1+x^8)| + C
∫(1-x^8)/[x(1+x^8)] dx
=-∫(1/x) dx + 2∫dx/[x(1+x^8)]
=-ln|x| +2∫dx/[x(1+x^8)]
let
x^4 = tany
4x^3 dx = (secy)^2 .dy
∫dx/[x(1+x^8)]
=(1/4)∫(4x^3 dx)/[x^4.(1+x^8)]
=(1/4)∫ (secy)^2 .dy/[tany(secy)^2]
=(1/4)∫ dy/tany
=(1/4)ln|siny| + C'
=(1/4)ln| x^4/√(1+x^8)| + C'
∫(1-x^8)/[x(1+x^8)] dx
=-ln|x| +2∫dx/[x(1+x^8)]
=-ln|x| +(1/2)ln| x^4/√(1+x^8)| + C
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