Q47 / 125 · BITSAT 2020 · Mathematics
$\int_{0}^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}+\mathrm{b}^{2}\right)}$ is
- A$\frac{\pi a b}{a+b}$
- B$\frac{\pi}{2(a+b)}$
- Correct answerC$\frac{\pi}{2 a b(a+b)}$
- D$\frac{\pi(a+b)}{2 a b}$
Solution
$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{1}{b^{2}-a^{2}}$
$\int_{0}^{\infty} \frac{\left(x^{2}+b^{2}\right)-\left(x^{2}+a^{2}\right)}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}$
$=\frac{1}{b^{2}-a^{2}} \int_{0}^{\infty}\left[\frac{1}{x^{2}+a^{2}}-\frac{1}{x^{2}+b^{2}}\right] d x$
$=\frac{1}{b^{2}-a^{2}}\left[\frac{1}{a} \tan ^{-1} \frac{x}{a}-\frac{1}{b} \tan ^{-1} \frac{x}{b}\right]_{0}^{\infty}$
$=\frac{1}{b^{2}-a^{2}}\left[\frac{\pi}{2 a}-\frac{\pi}{2 b}\right]=\frac{\pi}{2 a b(a+b)}$
Hence, the correct answer is option: 'C'.