1/(x^4+a^4)の不定積分

$$\displaystyle\int \frac{1}{x^4+a^4}dx=\frac{1}{4\sqrt{2}a^3} \log \frac{x^2+\sqrt{2}ax+a^2}{x^2-\sqrt{2}ax+a^2}+ \frac{1}{2\sqrt{2}a^2}\tan^{-1} \frac{\sqrt{2}ax}{a^2-x^2}+C$$















<証明>

分母を因数分解してから、部分分数分解します。$$\displaystyle\int \frac{1}{x^4+a^4}dx=\displaystyle\int \frac{1}{(x^2-\sqrt{2}ax+a^2)(x^2+\sqrt{2}ax+a^2)}dx$$
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。$$\frac{Ax+B}{x^2-\sqrt{2}ax+a^2}+\frac{Cx+D}{x^2+\sqrt{2}ax+a^2}=\frac{1}{(x^2-\sqrt{2}ax+a^2)(x^2+\sqrt{2}ax+a^2)} $$
両辺に \((x^2-\sqrt{2}ax+a^2)(x^2+\sqrt{2}ax+a^2)\) を掛けます。
\begin{alignat}{2}
&(x^2+\sqrt{2}ax+a^2)(Ax+B)+(x^2-\sqrt{2}ax+a^2)(Cx+D)=1\\
&\\
&Ax^3+\sqrt{2}Aax^2+Aa^2x+Bx^2+\sqrt{2}Bax+Ba^2\\
&    +Cx^3-\sqrt{2}Cax^2+Ca^2x+Dx^2-\sqrt{2}Dax+Da^2=1\\
&\\
&(A+C)x^3+(\sqrt{2}Aa+B-\sqrt{2}Ca+D)x^2\\
&    +(Aa^2+\sqrt{2}Ba+Ca^2-\sqrt{2}Da)x+(Ba^2+Da^2)=1\\
&
\end{alignat} 両辺の係数を比較します。

\(\displaystyle A+C=0, B+D=\frac{1}{a^2} \left( C=-A, D=\frac{1}{a^2}-B \right)\)
\begin{alignat}{2}
&\sqrt{2}Aa+B-\sqrt{2}Ca+D=0\\
&\sqrt{2}Aa+B+\sqrt{2}Aa+\frac{1}{a^2}-B=0\\
&2\sqrt{2}Aa=-\frac{1}{a^2}, A=-\frac{1}{2\sqrt{2}a^3}, C=\frac{1}{2\sqrt{2}a^3}
&\\
&\\
&Aa^2+\sqrt{2}Ba+Ca^2-\sqrt{2}Da=0\\
&Aa^2+\sqrt{2}Ba-Aa^2-\sqrt{2}a\left(\frac{1}{a^2}-B\right)=0\\
&2\sqrt{2}Ba=\frac{\sqrt{2}}{a}, B=\frac{1}{2a^2}, D=\frac{1}{2a^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+a^4}dx=\displaystyle\int \left(\frac{-\frac{1}{2\sqrt{2}a^3}x+\frac{1}{2a^2}}{x^2-\sqrt{2}ax+a^2}+\frac{\frac{1}{2\sqrt{2}a^3}x+\frac{1}{2a^2}}{x^2+\sqrt{2}ax+a^2} \right)dx\\
&=-\frac{1}{2\sqrt{2}a^3}\displaystyle\int \frac{x-\sqrt{2}a}{x^2-\sqrt{2}ax+a^2}dx+\frac{1}{2\sqrt{2}a^3}\displaystyle\int \frac{x+\sqrt{2}a}{x^2+\sqrt{2}ax+a^2}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}a}{x^2-\sqrt{2}ax+a^2}dx=\displaystyle\int \frac{\left(x-\frac{a}{\sqrt{2}}\right)-\frac{a}{\sqrt{2}}}{\left(x-\frac{a}{\sqrt{2}}\right)^2+\frac{a^2}{2}}dx\\
&=\displaystyle\int \left\{\frac{x-\frac{a}{\sqrt{2}}}{\left(x-\frac{a}{\sqrt{2}}\right)^2+\frac{a^2}{2}}-\frac{1}{\sqrt{2}}\cdot\frac{1}{\left(x-\frac{a}{\sqrt{2}}\right)^2+\frac{a^2}{2}}\right\}dx \\
&=\frac{1}{2} \log \left\{\left(x-\frac{a}{\sqrt{2}}\right)^2+\frac{a^2}{2}\right\}- \tan^{-1} \left\{\frac{\sqrt{2}}{a}\left(x-\frac{a}{\sqrt{2}}\right)\right\}\\
&=\frac{1}{2} \log (x^2-\sqrt{2}ax+a^2)- \tan^{-1} \left(\frac{\sqrt{2}}{a}x-1\right)
\end{alignat}

\begin{alignat}{2}
&(B) \displaystyle\int \frac{x+\sqrt{2}a}{x^2+\sqrt{2}ax+a^2}dx=\displaystyle\int \frac{\left(x+\frac{a}{\sqrt{2}}\right)+\frac{a}{\sqrt{2}}}{\left(x+\frac{a}{\sqrt{2}}\right)^2+\frac{a^2}{2}}dx\\
&=\displaystyle\int \left\{\frac{x+\frac{a}{\sqrt{2}}}{\left(x+\frac{a}{\sqrt{2}}\right)^2+\frac{a^2}{2}}+\frac{1}{\sqrt{2}}\cdot\frac{1}{\left(x+\frac{a}{\sqrt{2}}\right)^2+\frac{a^2}{2}}\right\}dx \\
&=\frac{1}{2} \log \left\{\left(x+\frac{a}{\sqrt{2}}\right)^2+\frac{a^2}{2}\right\}+ \tan^{-1} \left\{\frac{\sqrt{2}}{a}\left(x+\frac{a}{\sqrt{2}}\right)\right\}\\
&=\frac{1}{2} \log (x^2+\sqrt{2}ax+a^2)+ \tan^{-1} \left(\frac{\sqrt{2}}{a}x+1\right)
\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+a^4}dx=-\frac{1}{4\sqrt{2}a^3} \log (x^2-\sqrt{2}ax+a^2)+\frac{1}{2\sqrt{2}a^3} \tan^{-1} \left(\frac{\sqrt{2}}{a}x-1\right)\\
&               +\frac{1}{4\sqrt{2}a^3} \log (x^2+\sqrt{2}ax+a^2)+\frac{1}{2\sqrt{2}a^3} \tan^{-1} \left(\frac{\sqrt{2}}{a}x+1\right)\\
&        =\frac{1}{4\sqrt{2}a^3}\left\{ \log (x^2+\sqrt{2}ax+a^2)- \log (x^2-\sqrt{2}ax+a^2)\right\}\\
&               +\frac{1}{2\sqrt{2}a^3}\left\{ \tan^{-1}\left(\frac{\sqrt{2}}{a}x-1\right)+ \tan^{-1} \left(\frac{\sqrt{2}}{a}x+1\right)\right\}
\end{alignat}
\(\tan^{-1}\) の計算について
\begin{alignat}{2}
\tan^{-1}\left(\frac{\sqrt{2}}{a}x-1\right)+ \tan^{-1} \left(\frac{\sqrt{2}}{a}x+1\right)&=\tan^{-1} \frac{\frac{\sqrt{2}}{a}x-1+\frac{\sqrt{2}}{a}x+1}{1-\left(\frac{\sqrt{2}}{a}x-1\right)\left(\frac{\sqrt{2}}{a}x+1\right)}\\
&=\tan^{-1} \frac{\frac{2\sqrt{2}}{a}x}{1-\left(\frac{2}{a^2}x^2-1\right)}=\tan^{-1} \frac{\frac{2\sqrt{2}}{a}x}{2-\frac{2}{a^2}x^2}=\tan^{-1}\frac{\sqrt{2}ax}{a^2-x^2}\\
\end{alignat}以上より$$\displaystyle\int \frac{1}{x^4+a^4}dx=\frac{1}{4\sqrt{2}a^3} \log \frac{x^2+\sqrt{2}ax+a^2}{x^2-\sqrt{2}ax+a^2}+ \frac{1}{2\sqrt{2}a^2}\tan^{-1} \frac{\sqrt{2}ax}{a^2-x^2}+C$$

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