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Question: Answered & Verified by Expert
$\int_0^{\infty} \frac{x^2 d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}=$
MathematicsDefinite IntegrationJEE Main
Options:
  • A $\frac{\pi}{2(a-b)}$
  • B $\frac{\pi}{2(b-a)}$
  • C $\frac{\pi}{(a+b)}$
  • D $\frac{\pi}{2(a+b)}$
Solution:
1328 Upvotes Verified Answer
The correct answer is: $\frac{\pi}{2(a+b)}$
$\begin{aligned} & \int_0^{\infty} \frac{x^2 d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}=\int_0^{\infty} \frac{\left(x^2+a^2\right)-a^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)} d x \\ & \int_0^{\infty} \frac{1}{x^2+b^2} d x-a^2 \int_0^{\infty} \frac{1}{\left(x^2+a^2\right)\left(x^2+b^2\right)} d x \\ = & {\left[\frac{1}{b} \tan ^{-1} \frac{x}{b}\right]_0^{\infty}-\frac{a^2}{\left(a^2-b^2\right)} \int_0^{\infty}\left(\frac{1}{x^2+b^2}-\frac{1}{x^2+a^2}\right) d x } \\ = & \frac{1}{b} \cdot \frac{\pi}{2}-\frac{a^2}{\left(a^2-b^2\right)}\left[\frac{1}{b} \tan ^{-1} \frac{x}{b}-\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]_0^{\infty}=\frac{\pi}{2(a+b)}\end{aligned}$

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