The value of-0-dx-a2-x2- is

Explanation for the correct answer.Evaluate integral.∫_0^∞dx/[a^2+x^2]0ex0ex=1/a[tan^ 1x/a]_0^∞0ex0ex=1/a[π/2 0]0ex0ex=π/2aHence, option B is correct . ...

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Standard IX

Mathematics

Surds

The value of∫...

Question

The value of $\int_{0}^{\infty} \frac{d x}{[a^{2} + x^{2}]}$ is

$\frac{π}{2}$

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$\frac{π}{2 a}$

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$\frac{π}{a}$

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$\frac{1}{2 a}$

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Solution

The correct option is B
$\frac{π}{2 a}$

Explanation for the correct answer.
Evaluate integral.
$\int_{0}^{\infty} \frac{d x}{[a^{2} + x^{2}]} = \frac{1}{a} {[\tan^{- 1} \frac{x}{a}]}_{0}^{\infty} = \frac{1}{a} [\frac{π}{2} - 0] = \frac{π}{2 a}$
Hence, option B is correct .

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The value of∫0∞dxa2+x2 is

The correct option is B π2aExplanation for the correct answer.Evaluate integral.∫0∞dxa2+x2=1atan-1xa0∞=1aπ2-0=π2aHence, option B is correct .

The value of $\int_{0}^{\infty} \frac{d x}{[a^{2} + x^{2}]}$ is

The correct option is B
$\frac{π}{2 a}$

Explanation for the correct answer.
Evaluate integral.
$\int_{0}^{\infty} \frac{d x}{[a^{2} + x^{2}]} = \frac{1}{a} {[\tan^{- 1} \frac{x}{a}]}_{0}^{\infty} = \frac{1}{a} [\frac{π}{2} - 0] = \frac{π}{2 a}$
Hence, option B is correct .