日经题(3)

2020-01-31 本文已影响0人洛玖言

日经题(3)

$\displaystyle\int_0^1\dfrac{1}{1+x^4}\text{d}x$

这个题目也是日常在群里看到有人问的，其实也是比较明显的有理代数分式的积分，只不过拆分的方法比较特殊.

这边我们直接求原函数

第一种方法：
对分母进行分解
$\begin{aligned} &\dfrac{1}{1+x^4}\\ =&\dfrac{1}{x^4+2x^2+1-2x^2}\\ =&\dfrac{1}{(x^2+1)^2-(\sqrt{2}x)^2}\\ =&\dfrac{1}{(x^2-\sqrt{2}x+1) (x^2+\sqrt{2}x+1)} \end{aligned}$

到这里应该已经很明了了，只要用待定系数法分解就好了，不过还是把他补完整(毕竟疫情外面没人玩)

设
$\dfrac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}=\dfrac{Ax+B}{x^2-\sqrt{2}x+1}+\dfrac{Cx+D}{x^2+\sqrt{2}x+1}$

$(Ax+B)(x^2+\sqrt{2}x+1)+(Cx+D)(x^2-\sqrt{2}x+1)=1\\ (A+C)x^3+(\sqrt{2}A-\sqrt{2}C+B+D)x^2+(A+\sqrt{2}B+C-\sqrt{2}D)x+(B+D)=1\\ \Rightarrow \begin{cases} A+C=0\\ \sqrt{2}A-\sqrt{2}C+B+D=0\\ A+\sqrt{2}B+C-\sqrt{2}D=0\\ B+D=1 \end{cases}\\ \Rightarrow \begin{cases} A=-C=-\dfrac{\sqrt{2}}{4}\\ B=D=\dfrac12\\ \end{cases}$

$\begin{aligned} &\dfrac{1}{(x^2-\sqrt{2}x+1) (x^2+\sqrt{2}x+1)}\\ =&-\dfrac{2x-\sqrt{2}}{4\sqrt{2}(x^2-\sqrt{2}x+1)}+\dfrac{2x+\sqrt{2}}{4\sqrt{2}(x^2+\sqrt{2}x+1)}\\ &+\dfrac{\sqrt{2}}{4\sqrt{2}(x^2-\sqrt{2}x+1)}+\dfrac{\sqrt{2}}{4\sqrt{2}(x^2+\sqrt{2}x+1)} \end{aligned}$

$\begin{aligned} &\int\dfrac{1}{1+x^4}\text{d}x\\ =&\dfrac{1}{4\sqrt{2}}\left[\ln\left(\dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+2\arctan(\sqrt{2}x-1)+2\arctan(\sqrt{2}x+1)\right]+C \end{aligned}$

这种方法其实还是蛮麻烦的，但是蛮好想的
下面这种方法不用麻烦的待定系数法

解法二：
对分子进行处理
$\begin{aligned} &\dfrac{1}{1+x^4}\\ =&\frac{1}{2}\left(\dfrac{x^2+1}{1+x^4}-\dfrac{x^2-1}{1+x^4}\right) \end{aligned}$

$\begin{aligned} &\int\dfrac{1}{1+x^4}\text{d}x\\ =&\dfrac12\int\left(\dfrac{1+\frac1{x^2}}{x^2+\frac1{x^2}}-\dfrac{1-\frac1{x^2}}{x^2+\frac1{x^2}}\right)\text{d}x\\ =&\frac12\int\dfrac{\text{d}(x-\frac1x)}{(x-\frac1x)^2+2}-\frac12\int\dfrac{\text{d}(x+\frac1x)}{(x+\frac1x)^2-2}\\ =&\dfrac{\sqrt{2}}{4}\int\dfrac{\text{d}(\frac{x-\frac1x}{\sqrt{2}})}{(\frac{x-\frac1x}{\sqrt{2}})^2+1}-\frac12\int\dfrac{\text{d}(x+\frac1x)}{(x+\frac1x)^2-2}\\ =&\dfrac{\sqrt{2}}{4}\arctan\left(\dfrac{x-\frac1x}{\sqrt{2}}\right)+\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+C \end{aligned}$

补充：
$\arctan\alpha+\arctan\beta=\arctan\dfrac{\alpha+\beta}{1-\alpha\beta}$
$\arctan x+\arctan\dfrac1x=\dfrac\pi2$

日经题(3)

日经题(3)

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