MHT CET · Maths · Indefinite Integration
\(\int_{0}^{\infty} \frac{d x}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=\)
- A \(\frac{\pi}{120}\)
- B \(\frac{\pi}{60}\)
- C \(\frac{\pi}{80}\)
- D \(\frac{-\pi}{60}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{60}\)
Step-by-step Solution
Detailed explanation
Let \(x^{2}=t\)
\(\therefore \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=\frac{1}{(t+4)(t+9)} \text { and let }\) \(\frac{1}{(t+4)(t+9)}=\frac{1}{5}\left[\frac{1}{t+4}-\frac{1}{t+9}\right] \)
\( \therefore I =\frac{1}{5} \int_{0}^{\infty}\left[\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}\right] \mathrm{dx} \)
\( =\frac{1}{5}\left\{\left[\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right]_{0}^{\infty}-\left[\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)\right]_{0}^{\infty}\right\} \)
\( =\frac{1}{5}\left\{\frac{1}{2} \times \frac{\pi}{2}-\frac{1}{3} \times \frac{\pi}{2}\right\}=\frac{\pi}{10}\left\{\frac{1}{2}-\frac{1}{3}\right\}=\frac{\pi}{60}\)
\(\therefore \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)}=\frac{1}{(t+4)(t+9)} \text { and let }\) \(\frac{1}{(t+4)(t+9)}=\frac{1}{5}\left[\frac{1}{t+4}-\frac{1}{t+9}\right] \)
\( \therefore I =\frac{1}{5} \int_{0}^{\infty}\left[\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}\right] \mathrm{dx} \)
\( =\frac{1}{5}\left\{\left[\frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right]_{0}^{\infty}-\left[\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)\right]_{0}^{\infty}\right\} \)
\( =\frac{1}{5}\left\{\frac{1}{2} \times \frac{\pi}{2}-\frac{1}{3} \times \frac{\pi}{2}\right\}=\frac{\pi}{10}\left\{\frac{1}{2}-\frac{1}{3}\right\}=\frac{\pi}{60}\)
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