The partial fractions of $$x4+24x2+28x2+13$$ are

Correct option is 11x2+1+22x2+12+5x2+13

Questions

The partial fractions of $\frac{x^{4} + 24 x^{2} + 28}{{(x^{2} + 1)}^{3}}$ are

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1x2+1+22x2+12+5x2+13

1x2+1+22x2+12-5x2+1

1x2+1-22x2+12-5x2+13

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detailed solution

Correct option is A

x4+24x2+28x2+13=A1x+B1x2+1+A2x+B2x2+12+A3x+B3x2+13 ⇒x4+24x2+28=A1x+B1x2+12+A2x+B2x2+1+A3x+B3Putting x=i, 5=A3i+B3⇒A3=0, B3=5Equating different powers of x,0=A1B1=1, 2A1+A2=0⇒A2=0 2B1+B2=24⇒B2=22.∴Partial fraction =1x2+1+22x2+12+5x2+13.

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The partial fractions of x4+24x2+28x2+13 are

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The partial fractions of $\frac{x^{4} + 24 x^{2} + 28}{{(x^{2} + 1)}^{3}}$ are