<s>[INST] Integration consquence How does it follow from $$\int_0^\infty \left(\frac{1}{a^2+x^2}    \right) dx= \frac{\pi}{2a}$$ that $$\int_0^\infty \left(\frac{1}{\left({a^2+x^2}\right) ^2 }  \right) dx= \frac{\pi}{4a^3}\text{ ?}$$
 [/INST] From $\displaystyle\int_0^\infty \left(\frac{1}{a^2+x^2}    \right) dx= \frac{\pi}{2a}$, we can differentiate both sides with respect to $a$: 
$\displaystyle\int_0^\infty \dfrac{\partial}{\partial a}\left(\frac{1}{a^2+x^2}    \right) dx= \dfrac{\partial}{\partial a}\frac{\pi}{2a}$
$\displaystyle\int_0^\infty \left(-\frac{2a}{(a^2+x^2)^2}\right) dx= -\frac{\pi}{2a^2}$
Do you see how to finish? (Note you should also verify for yourself that the conditions for differentiating under the integral sign are met.)
EDIT: The other way to handle the second integral is to use trigonometric substitution. 
Specifically, let $x = a\tan \theta$. Then, $dx = a\sec^2\theta\,d\theta$ and $a^2+x^2 = a^2\sec^2\theta$. Hence, 
$\displaystyle\int_{0}^{\infty}\dfrac{dx}{(a^2+x^2)^2} = \int_{0}^{\pi/2}\dfrac{a\sec^2\theta\,d\theta}{(a^2\sec^2\theta)^2} = \dfrac{1}{a^3}\int_{0}^{\pi/2}\cos^2\theta\,d\theta$, which easily evaluates to $\dfrac{\pi}{4a^3}$. 
Of course, this doesn't "follow" from $\displaystyle\int_0^\infty \left(\frac{1}{a^2+x^2}    \right) dx= \frac{\pi}{2a}$, which is why everyone suggested to differentiate with respect to $a$.
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