Mathematics

$$\int\limits_0^\infty  {\dfrac{{dx}}{{({x^2} + {a^2})({x^2} + {b^2})}}}  = $$


SOLUTION
$$ \dfrac{1}{\left ( x^{2}+a^{2} \right )+\left ( x^{2}+b^{2} \right )} = \dfrac{1}{b^{2}-a^{2}}\left (  \dfrac{1}{\left ( x^{2}+a^{2} \right )}- \dfrac{1}{\left ( x^{2}+b^{2} \right )} \right )$$

$$So,\int_{0}^{\infty }\left ( \dfrac{1}{\left ( x^{2}+a^{2} \right )+\left ( x^{2}+b^{2} \right )} \right )dx= \dfrac{1}{b^{2}-a^{2}}\left (\int_{0}^{\infty }  \dfrac{1}{\left ( x^{2}+a^{2} \right )}dx \right )-\dfrac{1}{b^{2}-a^{2}}\left (\int_{0}^{\infty }  \dfrac{1}{\left ( x^{2}+b^{2} \right )}dx \right )$$

$$As \int \dfrac{1}{\left ( x^{2}+a^{2} \right )dx}=\\\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$$
$$So,\dfrac{1}{b^{2}-a^{2}}\left (\int_{0}^{\infty }  \dfrac{1}{\left ( x^{2}+a^{2} \right )}dx \right )-\dfrac{1}{b^{2}-a^{2}}\left (\int_{0}^{\infty }  \dfrac{1}{\left ( x^{2}+b^{2} \right )}dx \right ) = \\\\\dfrac{1}{b^{2}-a^{2}}\times \dfrac{1}{a}(\tan^{-1}\dfrac{x}{a})_{0}^{\infty } -\dfrac{1}{b^{2}-a^{2}} \times \dfrac{1}{b}(\tan^{-1}\dfrac{x}{b})_{0}^{\infty }$$
After putting limits , we will get
$$= \dfrac{\pi}{2(b^{2}-a^{2})}\left ( \dfrac{1}{a}-\dfrac{1}{b} \right )$$
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Subjective Medium Published on 17th 09, 2020
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