(1) $\frac{1}{x^4+1}$ を $\frac{Ax+B}{x^2+\sqrt{2}x+1} + \frac{Cx+D}{x^2-\sqrt{2}x+1}$ と部分分数分解したときの係数 $A$, $B$, $C$, $D$ を求めよ。 (2) $\int_0^1 \frac{dx}{x^4+1}$ の値を求めよ。

1. 問題の内容

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

(2) $\int_0^1 \frac{dx}{x^4+1}$ の値を求めよ。

2. 解き方の手順

(1) $\frac{1}{x^4+1} = \frac{Ax+B}{x^2+\sqrt{2}x+1} + \frac{Cx+D}{x^2-\sqrt{2}x+1}$ の両辺に $x^4+1 = (x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$ を掛けると、

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

$1 = Ax^3 - \sqrt{2}Ax^2 + Ax + Bx^2 - \sqrt{2}Bx + B + Cx^3 + \sqrt{2}Cx^2 + Cx + Dx^2 + \sqrt{2}Dx + D$

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

係数を比較すると、

$A+C = 0$

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

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

$B+D = 1$

$A+C=0$ より $C=-A$。

$B+D=1$ より $D=1-B$。

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

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

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

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

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

$\frac{\sqrt{2}}{4}-\sqrt{2}B-\frac{\sqrt{2}}{4}+\sqrt{2}(1-B) = 0$

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

$2\sqrt{2}B = \sqrt{2}$

$B = \frac{1}{2}$

$D = 1-\frac{1}{2} = \frac{1}{2}$

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

(2) $\int_0^1 \frac{dx}{x^4+1} = \int_0^1 \frac{\frac{\sqrt{2}}{4}x+\frac{1}{2}}{x^2+\sqrt{2}x+1} + \frac{-\frac{\sqrt{2}}{4}x+\frac{1}{2}}{x^2-\sqrt{2}x+1} dx$

$= \frac{\sqrt{2}}{4} \int_0^1 \frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1} + \frac{-x+\sqrt{2}}{x^2-\sqrt{2}x+1} dx$

$= \frac{\sqrt{2}}{4} \int_0^1 \frac{x+\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{x^2+\sqrt{2}x+1} + \frac{-(x-\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2}}{x^2-\sqrt{2}x+1} dx$

$= \frac{\sqrt{2}}{4} [\frac{1}{2} \log(x^2+\sqrt{2}x+1) + \sqrt{2} \arctan(\sqrt{2}x+1) - \frac{1}{2} \log(x^2-\sqrt{2}x+1) + \sqrt{2} \arctan(\sqrt{2}x-1)]_0^1$

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

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

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

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

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

$= \frac{\sqrt{2}}{4} [\frac{1}{2} \log((1+\sqrt{2})^2) + \frac{\sqrt{2}\pi}{2} ]$

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

$= \frac{\sqrt{2}}{4} \log(1+\sqrt{2}) + \frac{\pi}{4}$

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