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Question: Answered & Verified by Expert
$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=$
MathematicsIndefinite IntegrationMHT CETMHT CET 2020 (12 Oct Shift 1)
Options:
  • A $\frac{\pi}{120}$
  • B $\frac{\pi}{60}$
  • C $\frac{\pi}{80}$
  • D $\frac{-\pi}{60}$
Solution:
1224 Upvotes Verified Answer
The correct answer is: $\frac{\pi}{60}$
Let $x^{2}=t$
$\begin{aligned}
\therefore & \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=\frac{1}{(t+4)(t+9)} \text { and let } \frac{1}{(t+4)(t+9)}=\frac{1}{5}\left[\frac{1}{t+4}-\frac{1}{t+9}\right] \\
\therefore I &=\frac{1}{5} \int_{0}^{\infty}\left[\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}\right] \mathrm{dx} \\
&=\frac{1}{5}\left\{\left[\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right]_{0}^{\infty}-\left[\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)\right]_{0}^{\infty}\right\} \\
&=\frac{1}{5}\left\{\frac{1}{2} \times \frac{\pi}{2}-\frac{1}{3} \times \frac{\pi}{2}\right\}=\frac{\pi}{10}\left\{\frac{1}{2}-\frac{1}{3}\right\}=\frac{\pi}{60}
\end{aligned}$

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