Summing the integers.

First look at $y = \log_{5x}\dfrac{5}{x} \Longrightarrow (5x)^{y} = \dfrac{5}{x}$

$\Longrightarrow x^{y+1} = 5^{1 - y}$

$\Longrightarrow (y + 1)\log_{5}(x) = 1 - y$

$\Longrightarrow y(1 + \log_{5}(x)) = 1 - \log_{5}(x)$

$\Longrightarrow y = \dfrac{1 - \log_{5}(x)}{1 + \log_{5}(x)}.$

Now let $a = \log_{5}(x).$ Then the given equation becomes

$a^{2} + \dfrac{1 - a}{1 + a} = 1 \Longrightarrow a^{3} + a^{2} - a + 1 = 1 + a$

$\Longrightarrow a(a^{2} + a - 2) = 0 \Longrightarrow a(a + 2)(a - 1) = 0.$

So we have solutions $a = \log_{5}(x) = 0, 1, -2 \Longrightarrow x = 1, 5, \dfrac{1}{25}.$

The only integral values are $1$ and $5$ , the sum of which is $\boxed{6}.$