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I want to solve for the following $$\frac{1}{1+t^4} $$

I start by following

$$\frac{1}{1+t^4} = \frac{1}{(1 +t^2)^2 - (\sqrt{2}t)^2} = \frac{1}{(1+t^2- \sqrt{2}t)(1+t^2+\sqrt{2}t)} $$

$$\frac{1}{(1+t^2- \sqrt{2}t)(1+t^2+\sqrt{2}t)} = \frac{A}{1+t^2 + \sqrt{2}t} + \frac{B}{1+t^2 - \sqrt{2}t} $$

But the answers I get following above is incorrect. Can someone help me know how to solve such kind of partial fractions?

SJa
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  • The numerators $,A,B,$ in the last sum are not constants, but rather linear polynomials in $,t,$. – dxiv Apr 11 '18 at 04:40
  • When the denominator is quadratic, the numerator must be linear. Reread what your text has to say about quadratics in the denominator. – saulspatz Apr 11 '18 at 04:41
  • You cannot use Simple A and B here..The numerator of the quadtraic term here should be in the form of linear function of t – Edison Apr 11 '18 at 04:47

2 Answers2

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For quadtraic terms of denominator numerator must be in Linear Form with t's. So Correct form is Following

$$\frac{1}{(1+t^2- \sqrt{2}t)(1+t^2+\sqrt{2}t)} = \frac{At+B}{1+t^2 + \sqrt{2}t} + \frac{Ct+D}{1+t^2 - \sqrt{2}t} $$

Edison
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$\frac{1}{1+t^4} = \frac{1}{(1+t^2- \sqrt{2}t)(1+t^2+\sqrt{2}t)} = \frac{At+B}{1+t^2 + \sqrt{2}t} + \frac{Ct+D}{1+t^2 - \sqrt{2}t}$

$1 = (At+B)(1+t^2-\sqrt2t)+(Ct+D)(1+t^2+\sqrt2t)$

$1=At+At^3-\sqrt2At^2+B+Bt^2-\sqrt2Bt+Ct+Ct^3+\sqrt2Ct^2+D+Dt^2+\sqrt2Dt$

$1=t^3(A+C)+t^2(\sqrt2C-\sqrt2A-\sqrt2B+B+D)+t(A+C+\sqrt2D-\sqrt2B)+(B+D)$

comparing the coefficients gives;

$A+C=0\\\sqrt2C-\sqrt2A-\sqrt2B+B+D=0\\A+C+\sqrt2D-\sqrt2B=0\\B+D=1$

Solving the system of equations gives ,$A=\frac{2-\sqrt2}{4\sqrt2},B=\frac12,C=\frac{\sqrt2-2}{4\sqrt2},D=\frac12$

The Integrator
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