$\int_{- \infty}^{\infty} \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$ equals

\frac{π}{2 (a + b)}

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\frac{π}{a + b}

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\frac{π^{2}}{a + b}

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None of these

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Solution

The correct option is B $\frac{π}{a + b}$
$\int_{- \infty}^{\infty} \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$
We will resolve $\frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})}$ into partial fraction.
Let $x^{2} = y$
$\frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{y}{(y + a^{2}) (y + b^{2})}$
Now, $\frac{y}{(y + a^{2}) (y + b^{2})} = \frac{A}{(y + a^{2})} + \frac{B}{(y + b^{2})}$ .....(1)
$\Rightarrow y = A (y + b^{2}) + B (y + a^{2})$
When $y = - a^{2}, A = \frac{a^{2}}{a^{2} - b^{2}}$
When $y = - b^{2}, B = - \frac{b^{2}}{a^{2} - b^{2}}$
Put this value in (1)
$\frac{y}{(y + a^{2}) (y + b^{2})} = \frac{a^{2}}{(a^{2} - b^{2}) (y + a^{2})} - \frac{b^{2}}{(a^{2} - b^{2}) (y + b^{2})}$
Replacing $y$ by $x^{2}$
$\frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{a^{2}}{(a^{2} - b^{2}) (x^{2} + a^{2})} - \frac{b^{2}}{(a^{2} - b^{2}) (x^{2} + b^{2})}$
$\Rightarrow \int_{- \infty}^{\infty} \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x = \int_{- \infty}^{\infty} \frac{a^{2}}{(a^{2} - b^{2}) (x^{2} + a^{2})} d x - \int_{- \infty}^{\infty} \frac{b^{2}}{(a^{2} - b^{2}) (x^{2} + b^{2})} d x$
$= \frac{a}{(a^{2} - b^{2})} {[{tan}^{- 1} (\frac{x}{a})]}_{- \infty}^{\infty} - \frac{b}{(a^{2} - b^{2})} {[{tan}^{- 1} (\frac{x}{b})]}_{- \infty}^{\infty}$
$= \frac{a π}{(a^{2} - b^{2})} - \frac{b π}{(a^{2} - b^{2})}$
$= \frac{π}{(a + b)}$

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