被积函数有 4 次幂的时候，先尝试凑 2 次幂

一、题目

二、解析

由于被积函数 $\frac{1}{x^4+1}$ 的分母中含有 $x$ 的四次幂，所以，我们要在其分子中凑出来 $x$ 的二次幂，因为四次幂除以二次幂仍然可以得到二次幂，从而实现被积函数中次幂的统一。如果我们在分子中凑 $x$ 的三次幂，则就无法轻易实现被积函数中次幂的统一，所以：

$$\begin{array}{rl}I=& \int \frac{1}{x^4+1}\mathrm{d}x\\ \\ =& \frac{1}{2}\int \frac{(x^2+1)–\left(x^2–1\right)}{x^4+1}\mathrm{d}x\\ \\ =& \frac{1}{2}\int \frac{x^2+1}{x^4+1}\mathrm{d}x–\frac{1}{2}\int \frac{x^2–1}{x^4+1}\mathrm{d}x\\ \\ =& \frac{1}{2}\int \frac{\frac{x^2}{x^2}+\frac{1}{x^2}}{\frac{x^4}{x^2}+\frac{1}{x^2}}\mathrm{d}x–\frac{1}{2}\int \frac{\frac{x^2}{x^2}–\frac{1}{x^2}}{\frac{x^4}{x^2}+\frac{1}{x^2}}\mathrm{d}x\\ \\ =& \frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\mathrm{d}x–\frac{1}{2}\int \frac{1–\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\mathrm{d}x\\ \\ =& \frac{1}{2}\int \frac{\mathrm{d}(x-\frac{1}{x})}{{\left(x–\frac{1}{x}\right)}^2+2}–\frac{1}{2}\int \frac{\mathrm{d}(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2–2}\\ \\ \Rightarrow & \{\begin{array}{l}{t}_1=x–\frac{1}{x}\\ t_2=x+\frac{1}{x}\end{array}\\ \\ =& \frac{1}{2}\int \frac{1}{t_1^2+2}\mathrm{d}{t}_1–\frac{1}{2}\int \frac{1}{t_2^2–2}\mathrm{d}{t}_2\\ \\ =& \frac{1}{2}\int \frac{1}{2(\frac{t_1^2}{2}+1)}\mathrm{d}{t}_1–\frac{1}{2}\int \frac{1}{(t_2+\sqrt{2})(t_2–\sqrt{2})}\mathrm{d}{t}_2\\ \\ =& \frac{1}{4}\int \frac{1}{{\left(\frac{t_1}{\sqrt{2}}\right)}^2+1}\mathrm{d}{t}_1–\frac{1}{2}\cdot \frac{1}{2\sqrt{2}}\int \left(\frac{1}{t_2+\sqrt{2}}–\frac{1}{t_2–\sqrt{2}}\right)\mathrm{d}{t}_2\\ \\ =& \frac{\sqrt{2}}{4}\int \frac{1}{{\left(\frac{t_1}{\sqrt{2}}\right)}^2+1}\mathrm{d}\left(\frac{t_1}{\sqrt{2}}\right)+\frac{\sqrt{2}}{8}\int \left(\frac{1}{t_2+\sqrt{2}}–\frac{1}{t_2–\sqrt{2}}\right)\mathrm{d}{t}_2\\ \\ =& \frac{\sqrt{2}}{4}\arcsin \left(\frac{t_1}{\sqrt{2}}\right)+\frac{\sqrt{2}}{8}\ln \left|\frac{t_2+\sqrt{2}}{t_2–\sqrt{2}}\right|\\ \\ \Rightarrow & \{\begin{array}{l}{t}_1=x–\frac{1}{x}\\ t_2=x+\frac{1}{x}\end{array}\\ \\ =& \frac{\sqrt{2}}{4}\arcsin \frac{x–\frac{1}{x}}{\sqrt{2}}+\frac{\sqrt{2}}{8}\ln \left|\frac{x+\frac{1}{x}+\sqrt{2}}{x+\frac{1}{x}–\sqrt{2}}\right|\\ \\ =& \frac{\sqrt{\mathbf{2}}}{\mathbf{4}}\mathbf{arctan}\mathbf{}\frac{{\mathit{x}}^{\mathbf{2}}–\mathbf{1}}{\sqrt{\mathbf{2}}\mathit{x}}\mathbf{+}\frac{\sqrt{\mathbf{2}}}{\mathbf{8}}\mathbf{ln}\mathbf{}\left|\frac{{\mathit{x}}^{\mathbf{2}}\mathbf{+}\sqrt{\mathbf{2}}\mathit{x}\mathbf{+}\mathbf{1}}{{\mathit{x}}^{\mathbf{2}}–\sqrt{\mathbf{2}}\mathit{x}\mathbf{+}\mathbf{1}}\right|\mathbf{+}\mathit{C}\\ \\ =& \frac{\sqrt{\mathbf{2}}}{\mathbf{4}}\mathbf{arctan}\mathbf{}\frac{{\mathit{x}}^{\mathbf{2}}–\mathbf{1}}{\sqrt{\mathbf{2}}\mathit{x}}–\frac{\sqrt{\mathbf{2}}}{\mathbf{8}}\mathbf{ln}\mathbf{}\left|\frac{{\mathit{x}}^{\mathbf{2}}–\sqrt{\mathbf{2}}\mathit{x}\mathbf{+}\mathbf{1}}{{\mathit{x}}^{\mathbf{2}}\mathbf{+}\sqrt{\mathbf{2}}\mathit{x}\mathbf{+}\mathbf{1}}\right|\mathbf{+}\mathit{C}\end{array}$$

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