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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2020 · Mathematics · Definite Integration

∫_ 0 ^ ∞ dx ( ^ 2 + ^ 2 )( ^ 2 + ^ 2 ) is

∫0∞dx(\mathrmx2+\mathrma2)(\mathrmx2+\mathrmb2)\int_{0}^{\infty} \frac{\mathrm{dx}}{\left(\mathrmx^{2}+\mathrma^{2}\right)\left(\mathrmx^{2}+\mathrmb^{2}\right)} is

Options

  1. Aπaba+b\frac{\pi a b}{a+b}
  2. Bπ2(a+b)\frac{\pi}{2(a+b)}
  3. Cπ2ab(a+b)\frac{\pi}{2 a b(a+b)}
  4. Dπ(a+b)2ab\frac{\pi(a+b)}{2 a b}

Correct answer

C. π2ab(a+b)\frac{\pi}{2 a b(a+b)}

Step-by-step solution

∫0∞dx(x2+a2)(x2+b2)=1b2−a2∫0∞(x2+b2)−(x2+a2)(x2+a2)(x2+b2)=1b2−a2∫0∞[1x2+a2−1x2+b2]dx=1b2−a2[1atan⁡−1xa−1btan⁡−1xb]0∞=1b2−a2[π2a−π2b]=π2ab(a+b)\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{1}{b^{2}-a^{2}}\int_{0}^{\infty} \frac{\left(x^{2}+b^{2}\right)-\left(x^{2}+a^{2}\right)}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{1}{b^{2}-a^{2}} \int_{0}^{\infty}\left[\frac{1}{x^{2}+a^{2}}-\frac{1}{x^{2}+b^{2}}\right] d x=\frac{1}{b^{2}-a^{2}}\left[\frac{1}{a} \tan ^{-1} \frac{x}{a}-\frac{1}{b} \tan ^{-1} \frac{x}{b}\right]_{0}^{\infty}=\frac{1}{b^{2}-a^{2}}\left[\frac{\pi}{2 a}-\frac{\pi}{2 b}\right]=\frac{\pi}{2 a b(a+b)}
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