Q52 / 100 · BITSAT 2024 (Memory Based Paper 3) · Mathematics
$\int_0^{\infty} \frac{d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}$ is
- A$\frac{\pi a b}{a+b}$
- B$\frac{a b}{2(a+b)}$
- Correct answerC$\frac{\pi}{2 a b(a+b)}$
- D$\frac{\pi(a+b)}{2 a b}$
Solution
Let $\begin{aligned} I & =\int_0^{\infty} \frac{1}{\left(x^2+a^2\right)\left(x^2+b^2\right)} d x \\ I & =\frac{1}{a^2-b^2} \int_0^{\infty} \frac{\left(x^2+a^2\right)-\left(x^2+b^2\right)}{\left(x^2+a^2\right)\left(x^2+b^2\right)} d x \\ I & =\frac{1}{a^2-b^2} \int_0^{\infty} \frac{1}{\left(x^2+b^2\right)}-\frac{1}{\left(x^2+a^2\right)} d x \\ I & =\frac{1}{a^2-b^2}\left[\frac{1}{b} \tan ^{-1} \frac{x}{b}-\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]_0^{\infty} \\ I & =\frac{1}{a^2-b^2}\left[\frac{1}{b} \times \frac{\pi}{2}-\frac{1}{a} \times \frac{\pi}{2}\right] \\ I & =\frac{1}{(a+b)(a-b)}\left[\frac{a-b}{a b}\right] \times \frac{\pi}{2} \\ I & =\frac{\pi}{2 a b(a+b)}\end{aligned}$
Hence, the correct answer is option: 'C'.