4x2x4−1=4x2(x2−1)(x2+1)=4x2(x−1)(x+1)(x2+1)=ax−1+bx+1+cx+dx2+1\frac{4x^2}{x^4-1} = \frac{4x^2}{(x^2-1)(x^2+1)} = \frac{4x^2}{(x-1)(x+1)(x^2+1)} = \frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}x4−14x2=(x2−1)(x2+1)4x2=(x−1)(x+1)(x2+1)4x2=x−1a+x+1b+x2+1cx+d a(x+1)(x2+1)+b(x−1)(x2+1)+(cx+d)(x−1)(x+1)=4x2a(x+1)(x^2+1)+b(x-1)(x^2+1)+(cx+d)(x-1)(x+1)=4x^2a(x+1)(x2+1)+b(x−1)(x2+1)+(cx+d)(x−1)(x+1)=4x2 a(x3+x2+x+1)+b(x3−x2+x−1)+(cx3+dx2−cx−d)=4x2a(x^3+x^2+x+1)+b(x^3-x^2+x-1)+(cx^3+dx^2-cx-d)=4x^2a(x3+x2+x+1)+b(x3−x2+x−1)+(cx3+dx2−cx−d)=4x2 a+b+c=0a−b+d=4a+b−c=0a−b−d=0a+b+c=0 \qquad a-b+d=4 \qquad a+b-c=0 \qquad a-b-d=0a+b+c=0a−b+d=4a+b−c=0a−b−d=0 これから a=1a=1a=1 b=−1b=-1b=−1 c=0c=0c=0 d=2d=2d=2 が求まる よって 1x−1−1x+1+2x2+1\frac{1}{x-1}-\frac{1}{x+1}+\frac{2}{x^2+1}x−11−x+11+x2+12