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Given that ∞ ∫ 0 X 2 ( X 2 + a 2 ) ( X 2 + B 2 ) ( X 2 + C 2 ) D X = π 2 ( a + B ) ( B + C ) ( C + a ) , the Value of ∞ ∫ 0 D X ( X 2 + 4 ) ( X 2 + 9 ) , - Mathematics

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प्रश्न

Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]

विकल्प

  • \[\frac{\pi}{60}\]
  • \[\frac{\pi}{20}\]
  • \[\frac{\pi}{40}\]
  • \[\frac{\pi}{80}\]
MCQ

उत्तर

\[\frac{\pi}{60}\]

 

\[ \int_0^\infty \frac{1}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)} d x\]
\[ = \frac{1}{5} \int_0^\infty \frac{1}{\left( x^2 + 4 \right)} - \frac{1}{\left( x^2 + 9 \right)}dx\]
\[ = \frac{1}{5} \left[ \frac{1}{2} \tan^{- 1} \frac{x}{2} - \frac{1}{3} \tan^{- 1} \frac{x}{3} \right]_0^\infty \]
\[ = \frac{1}{5}\left[ \frac{1}{2} \times \frac{\pi}{2} - \frac{1}{3} \times \frac{\pi}{2} \right]\]
\[ = \frac{1}{5} \times \frac{\pi}{12}\]
\[ = \frac{\pi}{60}\]

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Definite Integrals
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अध्याय 20: Definite Integrals - MCQ [पृष्ठ ११८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
MCQ | Q 13 | पृष्ठ ११८

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