<s>[INST] Provide a different method of proving:$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}$ Accidentally founded this particular integral producing a rational number
I can't be for sure it is correct, so can one provide a proof of it. 
$$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}\tag
1$$
I found related to $(1)$ is this
Let enforce a substitution of $u=x+e^{\pi}$ then $du=dx$
$$\int_{-\infty}^{\infty}{1\over [\pi{u}^2+\pi^{1/3}]^2}du={1\over 2}\tag2$$
To avoid confusing with too much $\pi$ symbol, we write a general
$$\int_{-\infty}^{\infty}{1\over [A{u}^2+B]^2}du={1\over 2}\tag3$$
We could apply partial decomposition 
$${au+b\over Au^2+B}+{cu+d\over (Au^2+B)^2}=1\tag4$$ then find a,b,c and d.
I found a general integral of 
$$\int{dx\over(x^2+a^2)^2}={x\over2a^2(x^2+a^2)}+{1\over 2a^3}\tan^{-1}{\left(x\over a\right)}\tag5$$
 I am sure this is enough to prove  $(1)$

Question: what are other methods can we apply to prove $(1)?$

 [/INST] To avoid confusion, $a = A = \pi$ and $b = B = \pi ^{1/3}$
Let $z=\frac{a u^{2}}{b}$
\begin{align}
\int\limits_{-\infty}^{\infty} \frac{1}{(a u^{2} + b)^{2}} du &= 2 \int\limits_{0}^{\infty} \frac{1}{(a u^{2} + b)^{2}} du \\
&= \frac{2}{b^{2}} \int\limits_{0}^{\infty} \frac{1}{(\frac{a}{b} u^{2} + 1)^{2}} du \\
&= \frac{1}{b^{2}} \sqrt{\frac{b}{a}} \int\limits_{0}^{\infty} \frac{z^{-1/2}}{(z+1)^{2}} dz \\
&= a^{-1/2} b^{-3/2} \mathrm{B}\left(\frac{1}{2},\frac{3}{2} \right) \\
&= \frac{1}{\pi} \mathrm{B}\left(\frac{1}{2},\frac{3}{2} \right) \\
&= \frac{1}{\pi} \frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)} \\
&= \frac{1}{2}
\end{align}
Note that we used the following integral definition of the beta function
\begin{equation}
\mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{z^{x-1}}{(z+1)^{x+y}} dz
\end{equation}
</s>