Integration consquence

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 $a$.

HINT:

$$\frac{d}{da}\int_0^{\infty}\frac{1}{x^2+a^2}dx=-2a\int_0^{\infty}\frac{1}{(x^2+a^2)^2}dx$$

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

$\displaystyle\int_0^{\infty}\frac{1}{(x^2+a^2)^2}dx=\frac{1}{a^2}\left[\int_0^{\infty}\frac{1}{x^2+a^2}dx+\left[\frac{x}{2(x^2+a^2)}\right]_0^{\infty}-\frac{1}{2}\int_0^{\infty}\frac{1}{x^2+a^2}dx\right]$

$\displaystyle\hspace{1.3 in}=\frac{1}{a^2}\left[\frac{\pi}{2a}+0-\frac{1}{2}\cdot\frac{\pi}{2a}\right]=\frac{1}{a^2}\cdot\frac{\pi}{4a}=\frac{\pi}{4a^3}$.