Integration consquence How does it follow from $$\int_0^\infty \left(\frac{1}{a^2+x^2}    \right) dx= \frac{\pi}{2a}$$ that $$\int_0^\infty \left(\frac{1}{\left({a^2+x^2}\right) ^2 }  \right) dx= \frac{\pi}{4a^3}\text{ ?}$$
 A: 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 $a$.
A: Do what the question (STEP II 2014 Q4) says! Set $u=k/x$, then
$$ \int_0^{\infty} \frac{dx}{(a^2+x^2)^2} = \int_{\infty}^0 \frac{1}{(a^2+k^2/u^2)^2} \frac{-k du}{u^2} = \int_0^{\infty} \frac{ku^2 \, du}{(k^2+a^2 u^2)^2} $$
Now, use partial fractions.
$$ \frac{ku^2}{(k^2+a^2 u^2)^2} = \frac{k}{a^2}\frac{a^2 u^2+k^2-k^2}{(k^2+a^2 u^2)^2} = \frac{k}{a^2} \frac{1}{k^2+a^2 u^2} - \frac{k^3}{a^2} \frac{1}{(k^2+a^2 u^2)^2} $$
We know the integral of the first one. What we need to do now is choose $k$ so that the second ends up proportional to $1/(a^2+u^2)$. The obvious way is to take $k=a^2$, so that the above equation becomes
$$ \frac{u^2}{a^2(a^2+u^2)^2} = \frac{u^2+a^2-a^2}{a^2(a^2+u^2)^2} = \frac{1}{a^2} \frac{1}{a^2+u^2} - \frac{1}{(a^2+u^2)^2}. $$
So
$$ \int_0^{\infty} \frac{dx}{(a^2+x^2)^2} = \int_0^{\infty} \frac{u^2 \, du}{a^2(a^2+ u^2)^2} = \int_0^{\infty} \frac{1}{a^2} \frac{du}{a^2+u^2} - \int_0^{\infty} \frac{du}{(a^2 +u^2)^2}, $$
and the original integral has reappeared on the right. Rearranging gives
$$ \int_0^{\infty} \frac{dx}{(a^2+x^2)^2} = \frac{1}{2a^2} \int_0^{\infty} \frac{1}{a^2} \frac{du}{a^2+u^2} = \frac{\pi}{4a^3}. $$
A: HINT:
$$\frac{d}{da}\int_0^{\infty}\frac{1}{x^2+a^2}dx=-2a\int_0^{\infty}\frac{1}{(x^2+a^2)^2}dx$$
A: An alternate method would be to use integration by parts:
$\displaystyle\int_0^{\infty}\frac{1}{(x^2+a^2)^2}dx=\frac{1}{a^2}\left[\int_0^{\infty}\frac{1}{x^2+a^2}dx-\int_0^{\infty}\frac{x^2}{(x^2+a^2)^2}dx\right]$, 
so letting $\displaystyle u=x, \;dv=\frac{x}{(x^2+a^2)^2}dx, \;du=dx, \;v=-\frac{1}{2(x^2+a^2)}dx$ yields
$\displaystyle\int_0^{\infty}\frac{1}{(x^2+a^2)^2}dx=\frac{1}{a^2}\left[\int_0^{\infty}\frac{1}{x^2+a^2}dx+\left[\frac{x}{2(x^2+a^2)}\right]_0^{\infty}-\frac{1}{2}\int_0^{\infty}\frac{1}{x^2+a^2}dx\right]$
$\displaystyle\hspace{1.3 in}=\frac{1}{a^2}\left[\frac{\pi}{2a}+0-\frac{1}{2}\cdot\frac{\pi}{2a}\right]=\frac{1}{a^2}\cdot\frac{\pi}{4a}=\frac{\pi}{4a^3}$.
