Given that -0-x2 x2 - a2 x2 - b2 x2 - c2 dx - -2 a - b b - c c - a- the value of -0-dx x2 - 4 x2 - 9 is

The correct option is A π60 Given, ∫ _ 0 ^∞ #160; x^ 2 (x^ 2 +a^ 2 )(x^ 2 +b^ 2 )(x^ 2 +c^ 2 ) dx= π #160; 2(a+b)(b+c)(c+a) .....(1) ...

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Question

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

\frac{π}{60}

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\frac{π}{20}

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\frac{π}{40}

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\frac{π}{80}

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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{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}

f (x) = {sin}^{- 1} (\frac{x^{2} + 1}{x^{2} + 2})

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