(1) $\frac{1}{x^4+1}$ を $\frac{Ax+B}{x^2+\sqrt{2}x+1} + \frac{Cx+D}{x^2-\sqrt{2}x+1}$ のように部分分数分解したときの係数 $A, B, C, D$ を求める。 (2) 定積分 $\int_{0}^{1} \frac{dx}{x^4+1}$ の値を求める。

1. 問題の内容

(1) $\frac{1}{x^4+1}$ を $\frac{Ax+B}{x^2+\sqrt{2}x+1} + \frac{Cx+D}{x^2-\sqrt{2}x+1}$ のように部分分数分解したときの係数 $A, B, C, D$ を求める。

(2) 定積分 $\int_{0}^{1} \frac{dx}{x^4+1}$ の値を求める。

2. 解き方の手順

(1) 部分分数分解

与えられた式より、

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

右辺を展開して整理すると、

$1 = (A+C)x^3 + (-\sqrt{2}A+B+\sqrt{2}C+D)x^2 + (A-\sqrt{2}B+C+\sqrt{2}D)x + (B+D)$

係数を比較して、

$A+C=0$

$-\sqrt{2}A+B+\sqrt{2}C+D=0$

$A-\sqrt{2}B+C+\sqrt{2}D=0$

$B+D=1$

$A+C=0$ より $C=-A$。これを他の式に代入すると、

$-\sqrt{2}A+B-\sqrt{2}A+D=0$

$A-\sqrt{2}B-A+\sqrt{2}D=0$

$B+D=1$

整理して、

$-2\sqrt{2}A+B+D=0$

$-\sqrt{2}B+\sqrt{2}D=0$

$B+D=1$

$-\sqrt{2}B+\sqrt{2}D=0$ より $B=D$。$B+D=1$ より $B=D=\frac{1}{2}$。

$-2\sqrt{2}A + \frac{1}{2} + \frac{1}{2} = 0$

$-2\sqrt{2}A = -1$

$A = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

$C = -A = -\frac{\sqrt{2}}{4}$

よって、

$A=\frac{\sqrt{2}}{4}, B=\frac{1}{2}, C=-\frac{\sqrt{2}}{4}, D=\frac{1}{2}$

(2) 定積分の計算

(1)の結果を用いて積分を計算する。

$\int_{0}^{1} \frac{dx}{x^4+1} = \int_{0}^{1} \left( \frac{\frac{\sqrt{2}}{4}x+\frac{1}{2}}{x^2+\sqrt{2}x+1} + \frac{-\frac{\sqrt{2}}{4}x+\frac{1}{2}}{x^2-\sqrt{2}x+1} \right) dx$

$\int_{0}^{1} \frac{dx}{x^4+1} = \frac{\sqrt{2}}{4} \int_{0}^{1} \left( \frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1} + \frac{-x+\sqrt{2}}{x^2-\sqrt{2}x+1} \right) dx$

$\int_{0}^{1} \frac{dx}{x^4+1} = \frac{\sqrt{2}}{4} \int_{0}^{1} \left( \frac{x+\frac{\sqrt{2}}{2}}{x^2+\sqrt{2}x+1} + \frac{\frac{\sqrt{2}}{2}}{x^2+\sqrt{2}x+1} + \frac{-x+\frac{\sqrt{2}}{2}}{x^2-\sqrt{2}x+1} + \frac{\frac{\sqrt{2}}{2}}{x^2-\sqrt{2}x+1} \right) dx$

$\int_{0}^{1} \frac{dx}{x^4+1} = \frac{\sqrt{2}}{4} \left[ \frac{1}{2}\ln(x^2+\sqrt{2}x+1) - \frac{1}{2}\ln(x^2-\sqrt{2}x+1) + \arctan(\sqrt{2}x+1) + \arctan(\sqrt{2}x-1) \right]_{0}^{1}$

$=\frac{\sqrt{2}}{4}[\frac{1}{2} \ln(\frac{2+\sqrt{2}}{2-\sqrt{2}}) + arctan(1+\sqrt{2})-arctan(1-\sqrt{2})]$

$=\frac{\sqrt{2}}{4}[\frac{1}{2} \ln(\frac{(2+\sqrt{2})^2}{4-2}) + arctan(1+\sqrt{2})-arctan(1-\sqrt{2})]$

$=\frac{\sqrt{2}}{4}[\frac{1}{2} \ln(\frac{4+4\sqrt{2}+2}{2}) + arctan(1+\sqrt{2})-arctan(1-\sqrt{2})]$

$=\frac{\sqrt{2}}{4}[\frac{1}{2} \ln(3+2\sqrt{2}) + arctan(1+\sqrt{2})-arctan(1-\sqrt{2})]$

$=\frac{\sqrt{2}}{4}[\frac{1}{2} \ln((1+\sqrt{2})^2) + arctan(1+\sqrt{2})-arctan(1-\sqrt{2})]$

$=\frac{\sqrt{2}}{4}[\ln(1+\sqrt{2}) + arctan(1+\sqrt{2})-arctan(1-\sqrt{2})]$

arctan(1+sqrt(2))-arctan(1-sqrt(2)) = 3 pi/8

$= \frac{\sqrt{2}}{4}[\ln(1+\sqrt{2}) + \frac{3\pi}{8}]$

$=\frac{\sqrt{2}}{4}\ln(1+\sqrt{2})+\frac{3\sqrt{2}\pi}{32}$

3. 最終的な答え

(1) $A=\frac{\sqrt{2}}{4}, B=\frac{1}{2}, C=-\frac{\sqrt{2}}{4}, D=\frac{1}{2}$

(2) $\frac{\sqrt{2}}{4}\ln(1+\sqrt{2})+\frac{3\sqrt{2}\pi}{32}$

または$\frac{\pi\sqrt{2}+2\sqrt{2}\ln(1+\sqrt{2})}{8}$

$\frac{\sqrt{2}}{8}(\ln(1+\sqrt{2})+\frac{3\pi}{4})$

または$\frac{\sqrt{2}}{8}ln(3+2\sqrt{2}) + \frac{\sqrt{2}\pi}{4} \cdot \frac{3}{8} = \frac{\pi\sqrt{2}+2\sqrt{2}ln(1+\sqrt{2})}{8}$

最終答え：

（1）$A = \frac{\sqrt{2}}{4}, B = \frac{1}{2}, C = -\frac{\sqrt{2}}{4}, D = \frac{1}{2}$

（2）$\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{4}\ln(1+\sqrt{2})$

または、 $\frac{\sqrt{2}}{8}[\pi + 2 ln(1+\sqrt{2})]$