BITSAT 2024
Exam BITSAT
Year 2024
Subject Mathematics
Chapter Definite Integration BITSAT 2024 · Mathematics · Definite Integration
∫_0^ ∞ (d x)/((x^2+a^2)(x^2+b^2)) is
∫ 0 ∞ d x ( x 2 + a 2 ) ( x 2 + b 2 ) \int_0^{\infty} \frac{d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)} ∫ 0 ∞ ( x 2 + a 2 ) ( x 2 + b 2 ) d x is
Options A π a b a + b \frac{\pi a b}{a+b} a + b πab B a b 2 ( a + b ) \frac{a b}{2(a+b)} 2 ( a + b ) ab C π 2 a b ( a + b ) \frac{\pi}{2 a b(a+b)} 2 ab ( a + b ) π D π ( a + b ) 2 a b \frac{\pi(a+b)}{2 a b} 2 ab π ( a + b ) Correct answer C. π 2 a b ( a + b ) \frac{\pi}{2 a b(a+b)} 2 ab ( a + b ) π
Step-by-step solution
Let
I = ∫ 0 ∞ 1 ( x 2 + a 2 ) ( x 2 + b 2 ) d x I = 1 a 2 − b 2 ∫ 0 ∞ ( x 2 + a 2 ) − ( x 2 + b 2 ) ( x 2 + a 2 ) ( x 2 + b 2 ) d x I = 1 a 2 − b 2 ∫ 0 ∞ 1 ( x 2 + b 2 ) − 1 ( x 2 + a 2 ) d x I = 1 a 2 − b 2 [ 1 b tan − 1 x b − 1 a tan − 1 x a ] 0 ∞ I = 1 a 2 − b 2 [ 1 b × π 2 − 1 a × π 2 ] I = 1 ( a + b ) ( a − b ) [ a − b a b ] × π 2 I = π 2 a b ( a + b ) \begin{aligned} I & =\int_0^{\infty} \frac{1}{\left(x^2+a^2\right)\left(x^2+b^2\right)} d x \\ I & =\frac{1}{a^2-b^2} \int_0^{\infty} \frac{\left(x^2+a^2\right)-\left(x^2+b^2\right)}{\left(x^2+a^2\right)\left(x^2+b^2\right)} d x \\ I & =\frac{1}{a^2-b^2} \int_0^{\infty} \frac{1}{\left(x^2+b^2\right)}-\frac{1}{\left(x^2+a^2\right)} d x \\ I & =\frac{1}{a^2-b^2}\left[\frac{1}{b} \tan ^{-1} \frac{x}{b}-\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]_0^{\infty} \\ I & =\frac{1}{a^2-b^2}\left[\frac{1}{b} \times \frac{\pi}{2}-\frac{1}{a} \times \frac{\pi}{2}\right] \\ I & =\frac{1}{(a+b)(a-b)}\left[\frac{a-b}{a b}\right] \times \frac{\pi}{2} \\ I & =\frac{\pi}{2 a b(a+b)}\end{aligned} 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
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