Question

一道简单不定积分问题

Answer 1

分部积分
Jn=[u/(u^2+a^2)^n]-积分 u*d[(u^2+a^2)^(-n)]
=[u/(u^2+a^2)^n]+n 积分u*2u /(u^2+a^2)^(n+1)
=[u/(u^2+a^2)^n]+2n 积分(u^2)du/(u^2+a^2)^(n+1)
=[u/(u^2+a^2)^n]+2n 积分(u^2+a^2-a^2)du/(u^2+a^2)^(n+1)
=[u/(u^2+a^2)^n]+2n [ 积分du/(u^2+a^2)^n - a^2 积分du/(u^2+a^2)^(n+1)]
=[u/(u^2+a^2)^n]+2n [Jn-a^2 Jn+1]
=[u/(u^2+a^2)^n]+2n Jn - 2na^2 Jn+1
Jn+1=[(2n-1)Jn+u/(u^2+a^2)^n]/(2na^2)
用n-1替换n
Jn=[1/2a^2(n-1)][u/(u^2+a^2)^(n-1) + (2n-3)Jn-1]

不明白可追问

Answer 2