1/(x^4+1)の不定積分

$$\displaystyle\int \frac{1}{x^4+1}dx =\frac{1}{4\sqrt{2}} \log \frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}+ \frac{1}{2\sqrt{2}}\tan^{-1} \frac{\sqrt{2}x}{1-x^2}+C$$








<証明>

分母を因数分解してから、部分分数分解します。$$\displaystyle\int \frac{1}{x^4+1}dx=\displaystyle\int \frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}dx$$
次のように部分分数分解が出来たとして \(A,B,C,D\) の値を求めます。$$\frac{Ax+B}{x^2-\sqrt{2}x+1}+\frac{Cx+D}{x^2+\sqrt{2}x+1}=\frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)} $$
両辺に \((x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)\) を掛けます。
\begin{alignat}{2}
&(x^2+\sqrt{2}x+1)(Ax+B)+(x^2-\sqrt{2}x+1)(Cx+D)=1\\
&\\
&Ax^3+\sqrt{2}Ax^2+Ax+Bx^2+\sqrt{2}Bx+B\\
&    +Cx^3-\sqrt{2}Cx^2+Cx+Dx^2-\sqrt{2}Dx+D=1\\
&\\
&(A+C)x^3+(\sqrt{2}A+B-\sqrt{2}C+D)x^2\\
&    +(A+\sqrt{2}B+C-\sqrt{2}D)x+(B+D)=1\\
&
\end{alignat} 両辺の係数を比較します。

\(A+C=0, B+D=1 ( C=-A, D=1-B )\)
\begin{alignat}{2}
&\sqrt{2}A+B-\sqrt{2}C+D=0\\
&\sqrt{2}A+B+\sqrt{2}A+1-B=0\\
&2\sqrt{2}A=-1, A=-\frac{1}{2\sqrt{2}}, C=\frac{1}{2\sqrt{2}}
&\\
&\\
&A+\sqrt{2}B+C-\sqrt{2}D=0\\
&A+\sqrt{2}B-A-\sqrt{2}+\sqrt{2}B=0\\
&2\sqrt{2}B=\sqrt{2}, B=\frac{1}{2}, D=\frac{1}{2}
\end{alignat}

以上より、元の積分は次のようになります。
\begin{alignat}{2}
&\displaystyle\int \frac{1}{x^4+1}dx=\displaystyle\int \left(\frac{-\frac{1}{2\sqrt{2}}x+\frac{1}{2}}{x^2-\sqrt{2}x+1}+\frac{\frac{1}{2\sqrt{2}}x+\frac{1}{2}}{x^2+\sqrt{2}x+1} \right)dx\\
&=-\frac{1}{2\sqrt{2}}\displaystyle\int \frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}dx+\frac{1}{2\sqrt{2}}\displaystyle\int \frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1}dx
\end{alignat}
次に左右の積分を計算しますが、どちらも分母を平方完成して、

それに合わせて分子を整えます。そして分数を切り離して積分です。
\begin{alignat}{2}
&(A) \displaystyle\int \frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}dx=\displaystyle\int \frac{\left(x-\frac{1}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}}}{\left(x-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}}dx\\
&=\displaystyle\int \left\{\frac{x-\frac{1}{\sqrt{2}}}{\left(x-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}}-\frac{1}{\sqrt{2}}\cdot\frac{1}{\left(x-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}}\right\}dx \\
&=\frac{1}{2} \log \left\{\left(x-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}\right\}- \tan^{-1} \left\{\sqrt{2}\left(x-\frac{1}{\sqrt{2}}\right)\right\}\\
&=\frac{1}{2} \log (x^2-\sqrt{2}x+1)- \tan^{-1} (\sqrt{2}x-1)
\end{alignat}

\begin{alignat}{2}
&(B) \displaystyle\int \frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1}dx=\displaystyle\int \frac{\left(x+\frac{1}{\sqrt{2}}\right)+\frac{1}{\sqrt{2}}}{\left(x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}}dx\\
&=\displaystyle\int \left\{\frac{x+\frac{1}{\sqrt{2}}}{\left(x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}}+\frac{1}{\sqrt{2}}\cdot\frac{1}{\left(x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}}\right\}dx \\
&=\frac{1}{2} \log \left\{\left(x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}\right\}+ \tan^{-1} \left\{\sqrt{2}\left(x+\frac{1}{\sqrt{2}}\right)\right\}\\
&=\frac{1}{2} \log (x^2+\sqrt{2}x+1)+ \tan^{-1} (\sqrt{2}x+1)
\end{alignat}

以上より、積分の結果は次のようになります。
\begin{alignat}{2}
&\displaystyle\int \frac{1}{x^4+1}dx=-\frac{1}{4\sqrt{2}} \log (x^2-\sqrt{2}x+1)+\frac{1}{2\sqrt{2}} \tan^{-1} (\sqrt{2}x-1)\\
&               +\frac{1}{4\sqrt{2}} \log (x^2+\sqrt{2}x+1)+\frac{1}{2\sqrt{2}} \tan^{-1} (\sqrt{2}x+1)\\
&        =\frac{1}{4\sqrt{2}}\left\{ \log (x^2+\sqrt{2}x+1)- \log (x^2-\sqrt{2}x+1)\right\}\\
&               +\frac{1}{2\sqrt{2}}\left\{ \tan^{-1}(\sqrt{2}x-1)+ \tan^{-1} (\sqrt{2}x+1)\right\}
\end{alignat}\(\tan^{-1}\) の計算について$$\tan^{-1}(\sqrt{2}x-1)+\tan^{-1}(\sqrt{2}x+1)=\tan^{-1}\frac{\sqrt{2}x-1+(\sqrt{2}x+1)}{1-(\sqrt{2}x-1)(\sqrt{2}x+1)}=\tan^{-1}\frac{2\sqrt{2}x}{1-(2x^2-1)}=\tan^{-1} \frac{\sqrt{2}x}{1-x^2}$$となるので、以上より$$\displaystyle\int \frac{1}{x^4+1}dx =\frac{1}{4\sqrt{2}} \log \frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}+ \frac{1}{2\sqrt{2}}\tan^{-1} \frac{\sqrt{2}x}{1-x^2}+C$$

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