E of a signal using Rayleigh I have to find The energy of a signal using Rayleigh th. the signal is $$ x(t) = A e^{-At } u(t) $$ assuming A>0

Using the classic definition of E , I found that it should be $$ \frac{A}{2} $$

Using Rayleigh I should do 
$$  \int_{-inf}^{inf} | \frac{A}{A+i2\pi f}|^2 df $$
This because I previously found X(f)
From this I obtained 
$$\frac{A^2}{8 \pi^2  f}  log (\sqrt (A^2 +4\pi^2f^2)) $$ with log from -infinite to infinite. 

$$ log \sqrt x = log \frac{x}{2} $$ but now I have no idea how to continue , and if this is correct.. thank you
If I understand correctly, you want to verify the energy calculation in the frequency domain by computing the energy as

$$E_x=\int_{-\infty}^{\infty}|X(f)|^2df\tag{1}$$

with

$$X(f)=\mathcal{F}\big\{x(t)\big\}=\frac{A}{A+i2\pi f}\tag{2}$$

From $(2)$ we get

$$|X(f)|^2=\frac{A^2}{A^2+(2\pi f)^2}=\frac{1}{1+\left(\frac{2\pi f}{A}\right)^2}\tag{3}$$

With $(3)$ and with the substitution $x=2\pi f/A$, the integral $(1)$ becomes

$$E_x=\frac{A}{2\pi}\int_{-\infty}^{\infty}\frac{1}{1+x^2}dx=\frac{A}{2\pi}\arctan(x){\huge|}_{-\infty}^{\infty}=\frac{A}{2\pi}\cdot\pi=\frac{A}{2}\tag{4}$$