If -0-x2-x2-a2x2-b2x2-c2dx -2a-bb-cc-athen- -0-dx-x2-1x2-16-

∫_0^∞dx(x^2+4)(x^2+9) = ∫_0^∞x^2(x^2+4)(x^2+9)(x^2 + 0)dx nbsp; a=1,b=4, c=0 ⇒ nbsp; ∫_0^∞dx(x^2+4)(x^2+9) =π2(5)(4)(1)=π40 ...

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Infinite GP

If ∫0∞x2/x2+a...

Question

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

\frac{π}{2}

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\frac{π}{40}

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\frac{π}{80}

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\frac{π}{60}

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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)}

\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{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)},

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

\frac{π}{60}

\frac{π}{20}

\frac{π}{40}

\frac{π}{80}

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

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