Given that - 0 - x 2 - 2-a 2 x 2-b 2 x 2-c 2 d x- 2 a-b b-c c-a- the value of - 0 - d x x 2-4 x 2-9- isa - 60b - 20c - 40d - 80

(b) #160; #960;60 #8747;0 #8734;1x2+4x2+9dx=15 #8747;0 #8734;1x2+4 1x2+9dx=1512tan 1x2 13tan 1x30 #8734;=1512 #215; #960;2 13 #215; #960; ...

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Invertible Element Binary Operation

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Question

Given that $\int_{0}^{\infty} \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2}) (x^{2} + c^{2})} d x = \frac{π}{2 (a + b) (b + c) (c + a)},$ the value of $\int_{0}^{\infty} \frac{d x}{(x^{2} + 4) (x^{2} + 9)},$ is

(a) $\frac{π}{60}$

(b) $\frac{π}{20}$

(c) $\frac{π}{40}$

(d) $\frac{π}{80}$

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\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)}

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}^{\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)} =

\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{1}{(x^{2} + 4) (x^{2} + 9)} d x

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

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