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{aligned}
I = & \ \int \frac{1}{x^{4} + 1} \mathrm{~ d} x \\ \\
= & \ \frac{1}{2} \int \frac{\left(\textcolor{orangered}{ x^{2} } + 1 \right) – \left(\textcolor{orangered}{ 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}}{\textcolor{springgreen}{ x^{2} }} + \frac{1}{\textcolor{springgreen}{x^{2}}}}{\frac{x^{4}}{\textcolor{springgreen}{x^{2}}} + \frac{1}{\textcolor{springgreen}{x^{2}}}} \mathrm{~d} x – \frac{1}{2} \int \frac{\frac{x^{2}}{\textcolor{springgreen}{x^{2}}} – \frac{1}{\textcolor{springgreen}{x^{2}}}}{\frac{x^{4}}{\textcolor{springgreen}{x^{2}}} + \frac{1}{\textcolor{springgreen}{x^{2}}} } \mathrm{~d} x \\ \\
= & \ \frac{1}{2} \int \frac{\textcolor{pink}{1 + \frac{1}{x^{2}}}}{x^{2} + \frac{1}{x^{2} }} \mathrm{~d} x – \frac{1}{2} \int \frac{\textcolor{pink}{1 – \frac{1}{x^{2} }}}{x^{2} + \frac{1}{x^{2} }} \mathrm{~d} x \\ \\
= & \ \frac{1}{2} \int \frac{\textcolor{pink}{\mathrm{d} \left(x-\frac{1}{x} \right) }}{ \left(x – \frac{1}{x} \right)^{2} + 2 } – \frac{1}{2} \int \frac{\textcolor{pink}{\mathrm{d} \left ( x + \frac {1}{x} \right) }}{ \left( x + \frac{1}{x} \right)^{2} – 2 } \\ \\
\Rightarrow & \ \textcolor{gray}{\begin{cases}
t_{1} = x – \frac{1}{x} \\
t_{2} = x + \frac{1}{x}
\end{cases}} \\ \\
= & \ \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} – \textcolor{yellow}{ \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) + \textcolor{yellow}{\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 & \ \textcolor{gray}{\begin{cases}
t_{1} = x – \frac{1}{x} \\
t_{2} = x + \frac{1}{x}
\end{cases}} \\ \\
= & \ \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| \\ \\
= & \ \textcolor{springgreen}{\boldsymbol{ \frac{\sqrt{2}}{4} \arctan \frac{x^{2} – 1}{\sqrt{2} x } + \frac{\sqrt{2} }{8} \ln \left| \frac{x^{2} + \sqrt{2} x + 1 }{x^{2} – \sqrt{2} x + 1 } \right| + C }} \\ \\
= & \ \textcolor{springgreen}{\boldsymbol{ \frac{\sqrt{2}}{4} \arctan \frac{x^{2} – 1}{\sqrt{2} x } \ – \ \frac{\sqrt{2} }{8} \ln \left| \frac{x^{2} – \sqrt{2} x + 1 }{x^{2} + \sqrt{2} x + 1 } \right| + C }}
\end{aligned}
$$

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