∫ 0 ∞ ( x 2 + a 2 ) ( x 2 + b 2 ) d x is

Let I I I I I I I = ∫ 0 ∞ ( x 2 + a 2 ) ( x 2 + b 2 ) 1 d x = a 2 − b 2 1 ∫ 0 ∞ ( x 2 + a 2 ) ( x 2 + b 2 ) ( x 2 + a 2 ) − ( x 2 + b 2 ) d x = a 2 − b 2 1 ∫ 0 ∞ ( x 2 + b 2 ) 1 − ( x 2 + a 2 ) 1 d x = a 2 − b 2 1 [ b 1 tan − 1 b x − a 1 tan − 1 a x ] 0 ∞ = a 2 − b 2 1 [ b 1 × 2 π − a 1 × 2 π ] = ( a + b ) ( a − b ) 1 [ ab a − b ] × 2 π = 2 ab ( a + b ) π

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationMiscellaneous/MixedEasy2 minPYQ_2024

MathematicsEasysingle choice

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

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Answer:

Solution:

Let $I I I I I I I = \int_{0}^{\infty} \frac{1}{( x ^{2} + a ^{2} ) ( x ^{2} + b ^{2} )} d x = \frac{1}{a ^{2} - b ^{2}} \int_{0}^{\infty} \frac{( x ^{2} + a ^{2} ) - ( x ^{2} + b ^{2} )}{( x ^{2} + a ^{2} ) ( x ^{2} + b ^{2} )} d x = \frac{1}{a ^{2} - b ^{2}} \int_{0}^{\infty} \frac{1}{( x ^{2} + b ^{2} )} - \frac{1}{( x ^{2} + a ^{2} )} d x = \frac{1}{a ^{2} - b ^{2}} [\frac{1}{b} tan^{- 1} \frac{x}{b} - \frac{1}{a} tan^{- 1} \frac{x}{a}]_{0}^{\infty} = \frac{1}{a ^{2} - b ^{2}} [\frac{1}{b} \times \frac{π}{2} - \frac{1}{a} \times \frac{π}{2}] = \frac{1}{( a + b ) ( a - b )} [\frac{a - b}{ab}] \times \frac{π}{2} = \frac{π}{2 ab ( a + b )}$

Stream:BITSATSubject:MathematicsTopic:Definite IntegrationSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2024

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