The correct option is A π/2ab(a+b) I=∫ _ 0 ^∞ nbsp; dx ( x ^ 2 + a ^ 2 ) ( x ^ 2 + b ^ 2 ) nbsp; nbsp; = 1 ( a ^ 2 b^ 2 ) nbsp;∫ _ 0 ...

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Integration of Piecewise Continuous Functions

∫0∞dx/x2+a2x2...

Question

$\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} =$

\frac{π a b}{(a + b)}

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\frac{π}{2 (a + b)}

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\frac{π}{2 a b (a + b)}

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\frac{π (a + b)}{a b}

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Solution

The correct option is A $\frac{π}{2 a b (a + b)}$
$I = \int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{1}{(a^{2} - b^{2})} \int_{0}^{\infty} \frac{(x^{2} + a^{2}) - (x^{2} + b^{2})}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x = \frac{1}{(a^{2} - b^{2})} \int_{0}^{\infty} ⎛ ⎜ ⎝ \frac{1}{(x^{2} + b^{2})} - \frac{1}{(x^{2} + a^{2})} ⎞ ⎟ ⎠ d x = \frac{1}{(a^{2} - b^{2})} ({[\frac{1}{b} {t a n}^{- 1} \frac{x}{b}]}_{0}^{\infty} - {[\frac{1}{a} {t a n}^{- 1} \frac{x}{a}]}_{0}^{\infty}) = \frac{1}{(a^{2} - b^{2})} (\frac{π}{2 b} - \frac{π}{2 a}) = \frac{π}{2 a b (a + b)}$

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Similar questions

If \infty \int 0 \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2}) (x^{2} + c^{2})} d x = \frac{π}{2 (a + b) (b + c) (c + a)} then, \infty \int 0 \frac{d x}{(x^{2} + 1) (x^{2} + 16)} =

\int_{0}^{\frac{π}{2}} \frac{d x}{a^{2} {cos}^{2} x + b^{2} {sin}^{2} x} = \frac{π}{2 a b} (a, b > 0)

\int_{0}^{\infty} \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2}) (x^{2} + c^{2})} =

\frac{π}{2 (a + b) (b + c) (c + a)}

\int_{0}^{\infty} \frac{d x}{(x^{2} + 4) (x^{2} + 9)} =

\infty \int 0 \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2}) (x^{2} + c^{2})} d x = \frac{π}{2 (a + b) (b + c) (c + a)}

\infty \int 0 \frac{d x}{(x^{2} + 4) (x^{2} + 9)}

a, b > 0

\infty \int 0 \frac{ln (b x)}{x^{2} + a^{2}} d x = \frac{π}{2 a},

a b

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