题目

∫dx/(x^2+a^)^n=x/[2(n-1)a^2(x^2+a^2)^n-1]+(2n-3)/[2(n-1)a^2]∫dx/(x^2+a^)^n-1,

提问时间：2021-01-01

答案

设S(n) = ∫ dx/(x² + a²)ⁿ
S(n) = x/(x² + a²)ⁿ - ∫ x d[1/(x² + a²)ⁿ]
= x/(x² + a²)ⁿ - ∫ x * (-2nx) * 1/(a² + x²)ⁿ⁺¹ dx
= x/(x² + a²)ⁿ + 2n∫ x²/(a² + x²)ⁿ⁺¹ dx
= x/(x² + a²)ⁿ + 2n∫ (x² + a² - a²)/(a² + x²)ⁿ⁺¹ dx
= x/(x² + a²)ⁿ + 2n * S(n) - 2na² * S(n + 1)
==> 2na² * S(n + 1) = x/(x² + a²)ⁿ + (2n - 1) * S(n)
==> S(n + 1) = x/[2na²(x² + a²)ⁿ] + (2n - 1)/(2na²) * S(n)
==> S(n) = x/[2(n - 1)a²(x² + a²)ⁿ⁻¹] + (2n - 3)/[2(n - 1)a²] * S(n - 1)
===>
∫ dx/(x² + a²)ⁿ = x/[2(n - 1)a²(x² + a²)ⁿ⁻¹] + (2n - 3)/[2(n - 1)a²] * ∫ dx/(x² + a²)ⁿ⁻¹
分部积分法就是有这个好处,往往在推导Reduction Formula时用得上的.