Question

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

Easy

A

$\frac{1}{(x^{2} + 1)} + \frac{22}{{(x^{2} + 1)}^{2}} + \frac{5}{{(x^{2} + 1)}^{3}}$
B

$\frac{1}{(x^{2} + 1)} + \frac{22}{{(x^{2} + 1)}^{2}} - \frac{5}{(x^{2} + 1)}$
C

$\frac{1}{(x^{2} + 1)} - \frac{22}{{(x^{2} + 1)}^{2}} - \frac{5}{{(x^{2} + 1)}^{3}}$
D

None of these

Solution

$\frac{x^{4} + 24 x^{2} + 28}{{(x^{2} + 1)}^{3}} = \frac{A_{1} x + B_{1}}{x^{2} + 1} + \frac{A_{2} x + B_{2}}{{(x^{2} + 1)}^{2}} + \frac{A_{3} x + B_{3}}{{(x^{2} + 1)}^{3}} \Rightarrow x^{4} + 24 x^{2} + 28 = (A_{1} x + B_{1}) {(x^{2} + 1)}^{2} + (A_{2} x + B_{2}) (x^{2} + 1) + (A_{3} x + B_{3})$
Putting $x = i, 5 = A_{3} i + B_{3} \Rightarrow A_{3} = 0, B_{3} = 5$
Equating different powers of x,
$0 = A_{1} B_{1} = 1, 2 A_{1} + A_{2} = 0 \Rightarrow A_{2} = 0 2 B_{1} + B_{2} = 24 \Rightarrow B_{2} = 22 .$
$∴$ Partial fraction $= \frac{1}{x^{2} + 1} + \frac{22}{{(x^{2} + 1)}^{2}} + \frac{5}{{(x^{2} + 1)}^{3}} .$

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