Integration consquence How does it follow from $$\int_0^\infty \left(\frac{1}{a^2+x^2}    \right) dx= \frac{\pi}{2a}$$ that $$\int_0^\infty \left(\frac{1}{\left({a^2+x^2}\right) ^2 }  \right) dx= \frac{\pi}{4a^3}\text{ ?}$$

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
$$ \int_0^{\infty} \frac{dx}{(a^2+x^2)^2} = \frac{1}{2a^2} \int_0^{\infty} \frac{1}{a^2} \frac{du}{a^2+u^2} = \frac{\pi}{4a^3}. $$
