梗图上这个积分怎么算？

2022-02-07 14:42 作者:湮灭的末影狐 0人读过 | 我要投稿

前几天一个群里聊到这张梗图

这个积分还真是挺不容易的。

首先，我们可以尝试将分母因式分解。首先显然

$x%5E5%2B1%3D0%5CRightarrow%20x%3D0%2Ce%5E%7B%5Cpm%20i%5Cpi%2F5%7D%2Ce%5E%7B%5Cpm%203i%5Cpi%2F5%7D$

所幸36°和72°的三角函数都不是超越数，所以

$x%5E5%2B1%20%3D%20(x%2B1)(x%5E2-%5Cfrac%7B%5Csqrt%205%20%2B1%7D%7B2%7Dx%2B1)(x%5E2%2B%5Cfrac%7B%5Csqrt%205%20-1%7D%7B2%7Dx%2B1)$

由于我们原则上利用换元积分已经解决了以下两个积分：

$%5Cint%5Cfrac%7B%5Cmathrm%20d%20x%7D%7Bx%2Ba%7D%20%3D%20%5Cln%7Cx%2Ba%7C%2BC$

$%5Cint%20%5Cfrac%7Bk%2Bx%7D%7Ba%20x%5E2%2Bb%20x%2Bc%7D%5Cmathrm%20d%20x%20%3D%20%20%5Cfrac1%7B2a%7D%5Cln%20(a%20x%5E2%2Bb%20x%2Bc)-%5Cfrac%7B%20(b-2%20a%20k)%20%5Carctan%5Cleft(%5Cfrac%7B2%20a%20x%2Bb%7D%7B%5Csqrt%7B4%20a%20c-b%5E2%7D%7D%5Cright)%7D%7Ba%5Csqrt%7B4%20a%20c-b%5E2%7D%7D$

所以我们尝试将被积函数按下式分解：

$%5Cbegin%7Baligned%7D%0A%5Cfrac1%7Bx%5E5%2B1%7D%26%3D%5Cfrac1%7B%20(x%2B1)(x%5E2-%5Cfrac%7B%5Csqrt%205%20%2B1%7D%7B2%7Dx%2B1)(x%5E2%2B%5Cfrac%7B%5Csqrt%205%20-1%7D%7B2%7Dx%2B1)%7D%5C%5C%0A%26%3D%5Cfrac%7BA%7D%7Bx%2B1%7D%2B%5Cfrac%7BBx%2BC%7D%7Bx%5E2-%5Cfrac%7B%5Csqrt%205%20%2B1%7D%7B2%7Dx%2B1%7D%2B%5Cfrac%7BDx%2BE%7D%7Bx%5E2%2B%5Cfrac%7B%5Csqrt%205%20-1%7D%7B2%7Dx%2B1%7D%0A%5Cend%7Baligned%7D$

其中 ABCDE 都是待定系数。

由此，

$A(x%5E2-%5Cfrac%7B%5Csqrt%205%20%2B1%7D%7B2%7Dx%2B1)(x%5E2%2B%5Cfrac%7B%5Csqrt%205%20-1%7D%7B2%7Dx%2B1)%2B(Bx%2BC)(x%2B1)(x%5E2%2B%5Cfrac%7B%5Csqrt%205%20-1%7D%7B2%7Dx%2B1)%2B(Dx%2BE)(x%2B1)(x%5E2-%5Cfrac%7B%5Csqrt%205%20%2B1%7D%7B2%7Dx%2B1)%3D1$