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 $(1)?$


Starting with,
$$\int_{-\infty}^{\infty}{1\over [A{u}^2+B]^2}du$$ you could let $u=\sqrt{\frac{B}{A}} \tan(\theta)$ so $du=\sqrt{\frac{B}{A}} \sec^2(\theta)d \theta.$  Using $1+\tan^2(x)=\sec^2(x)$, this gives, $$\sqrt{\frac{B}{A}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2(\theta)}{(B\tan^2(\theta)+B)^2}d\theta=\sqrt{\frac{1}{AB^3}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta)d\theta.$$  Now using $cos^2(x)=\frac{1}{2}(cos(2x)+1)$,  we have, $$\frac{1}{2\sqrt{AB^3}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(2\theta)+1 d\theta=\frac{1}{2\sqrt{AB^3}}(\frac{\sin(2\theta)}{2} +\theta)|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{2\sqrt{AB^3}}.$$  Plugging in $A=\pi$ and $B=\pi^{\frac{1}{3}}$ gives the desired result.
