∫dx/(x^2+a^)^n=x/[2(n-1)a^2(x^2+a^2)^n-1]+(2n-3)/[2(n-1)a^2]
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∫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,
![∫dx/(x^2+a^)^n=x/[2(n-1)a^2(x^2+a^2)^n-1]+(2n-3)/[2(n-1)a^2]](/uploads/image/z/20296015-7-5.jpg?t=%E2%88%ABdx%2F%28x%5E2%2Ba%5E%29%5En%3Dx%2F%5B2%28n-1%29a%5E2%28x%5E2%2Ba%5E2%29%5En-1%5D%2B%282n-3%29%2F%5B2%28n-1%29a%5E2%5D)
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.
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.
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