If frac{1}{x^4+1}=frac{A x+B}{x^2+sqrt{2} x+1 | vedclass

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(A) Given that $\frac{1}{x^4+1}=\frac{A x+B}{x^2+\sqrt{2} x+1}+\frac{C x+D}{x^2-\sqrt{2} x+1}$.Multiplying both sides by $(x^4+1)$,we get $(A x+B)(x^2

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$\frac{3}{8}$

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(A) Given that $\frac{1}{x^4+1}=\frac{A x+B}{x^2+\sqrt{2} x+1}+\frac{C x+D}{x^2-\sqrt{2} x+1}$.Multiplying both sides by $(x^4+1)$,we get $(A x+B)(x^2-\sqrt{2} x+1)+(C x+D)(x^2+\sqrt{2} x+1)=1$.Comparing coefficients:For $x^3$: $A+C=0 \Rightarrow C=-A$.For $x^0$ (constant term): $B+D=1$.For $x^2$: $B-\sqrt{2} A+D+\sqrt{2} C=0 \Rightarrow (B+D)-\sqrt{2}(A-C)=0$.Since $B+D=1$ and $C=-A$,we have $1-\sqrt{2}(2A)=0 \Rightarrow A=\frac{1}{2\sqrt{2}}$ and $C=-\frac{1}{2\sqrt{2}}$.For $x^1$: $A-\sqrt{2} B+C+\sqrt{2} D=0 \Rightarrow (A+C)+\sqrt{2}(D-B)=0$.Since $A+C=0$,we have $\sqrt{2}(D-B)=0 \Rightarrow D=B$.Since $B+D=1$,we get $2B=1 \Rightarrow B=\frac{1}{2}$ and $D=\frac{1}{2}$.Now,$B D-A C = (\frac{1}{2})(\frac{1}{2}) - (\frac{1}{2\sqrt{2}})(-\frac{1}{2\sqrt{2}}) = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$.

If $\frac{1}{x^4+1}=\frac{A x+B}{x^2+\sqrt{2} x+1}+\frac{C x+D}{x^2-\sqrt{2} x+1}$ then $B D-A C=$

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