Question

(1-x^8)/x(1+x^8)的不定积分

求各位大神

Answer 1

∫(1-x^8)/x(1+x^8)dx
=∫(1-x^8)/(x+x^9)dx
=∫(-1/9-x^8+10/9)/(x+x^9)dx
=(10/9)*∫dx/(x+x^9)-(1/9)*∫d(x+x^9)/(x+x^9)
∫dx/(x+x^9) 令x=1/t
=∫(-1/t^2)/(1/t+1/t^9)dt
=-∫(t^7)/(t^8+1)dt
=-(1/8)*∫d(t^8+1)/(t^8+1)
=-(1/8)*ln|t^8+1|+C
=-(1/8)*ln|1/x^8+1|+C
∫d(x+x^9)/(x+x^9)
=ln|x+x^9|+C
所以原式=-(5/36)*ln|1/x^8+1|-(1/9)*ln|x+x^9|+C，其中C是任意常数

Answer 2

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

Answer 3

详情如图所示

有任何疑惑，欢迎追问