Given that ^ _0 x^2 (x^2+a^2)(x^2+b^2)(x^2+c^2) dx= 2(a+b)(b+c)(c… - Vidyadip

Question

Given that $\int\limits^{\infty}_0\frac{\text{x}^2}{(\text{x}^2+\text{a}^2)(\text{x}^2+\text{b}^2)(\text{x}^2+\text{c}^2)}\text{ dx}=\frac{\pi}{2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})},$ the value of $\int\limits^\infty_0\frac{1}{(\text{x}^2+4)(\text{x}^2+9)},$ is:

$\frac{\pi}{60}$
$\frac{\pi}{20}$
$\frac{\pi}{40}$
$\frac{\pi}{80}$

✓

Answer

$\frac{\pi}{60}$

Solution:

$\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)\big(\text{x}^2+9\big)}\text{dx}$

$=\frac{1}{5}\int\limits^\infty_0\frac{1}{\big(\text{x}^2+4\big)}-\frac{1}{\big(\text{x}^2+9\big)}\text{dx}$

$=\frac{1}{5}\bigg[\frac{1}{2\tan^{-1}{}}\frac{\text{x}}{2}-\frac{1}{3}\tan^{-1}\frac{\text{x}}{3}\bigg]^\infty_0$

$=\frac{1}{5}\bigg[\frac{1}{2}\times\frac{\pi}{2}-\frac{1}{3}\times\frac{\pi}{2}\bigg]$

$=\frac{1}{5}\times\frac{\pi}{12}$

$=\frac{\pi}{60}$

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