高等数学，下图第二个式子怎么展开因式？求稍微详细点的步骤
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(1) 或 1/[x(x+2)^2]=a/x+b/(x+1)+c/(x+1)^2
(2) (x^3-3x)/(x^2-1)^2= a/(x+1)+b/(x+1)^2+c/(x-1)+d/(x-1)^2
或 (x^3-3x)/(x^2-1)^2 =（A+Bx)/(x+1)^2+（C+Dx)/(x-1)^2
结果是 (x^3-3x)/(x^2-1)^2 = (1/2)[1/(x+1)+1/(x+1)^2+1/(x-1)-1/(x-1)^2]
(2) (x^3-3x)/(x^2-1)^2= a/(x+1)+b/(x+1)^2+c/(x-1)+d/(x-1)^2
或 (x^3-3x)/(x^2-1)^2 =（A+Bx)/(x+1)^2+（C+Dx)/(x-1)^2
结果是 (x^3-3x)/(x^2-1)^2 = (1/2)[1/(x+1)+1/(x+1)^2+1/(x-1)-1/(x-1)^2]
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