Solution Note: x4+2x2+1=(x2+1)2x^4+2x^2+1=(x^2+1)^2x4+2x2+1=(x2+1)2 So: x4x4+2x2+1=x4(x2+1)2\frac{x^4}{x^4+2x^2+1}=\frac{x^4}{(x^2+1)^2}x4+2x2+1x4=(x2+1)2x4 Divide first: x4(x2+1)2=1+−2x2−1(x2+1)2\frac{x^4}{(x^2+1)^2}=1+\frac{-2x^2-1}{(x^2+1)^2}(x2+1)2x4=1+(x2+1)2−2x2−1 Now write: −2x2−1(x2+1)2=Ax+Bx2+1+Cx+D(x2+1)2\frac{-2x^2-1}{(x^2+1)^2}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}(x2+1)2−2x2−1=x2+1Ax+B+(x2+1)2Cx+D Solving gives A=0A=0A=0, B=−2B=-2B=−2, C=0C=0C=0, D=1D=1D=1. Hence: x4x4+2x2+1=1−2x2+1+1(x2+1)2\boxed{\frac{x^4}{x^4+2x^2+1}=1-\frac{2}{x^2+1}+\frac{1}{(x^2+1)^2}}x4+2x2+1x4=1−x2+12+(x2+1)21