Will Euler-Mascheroni help ?

First let us substitute $\displaystyle x^{128} = u$

So we get $\displaystyle \frac{1}{128}\int_{0}^{\infty}\left(\frac{(e^{-u}-\frac{1}{1+u^{2}})}{u}\right)du$

Let us forget about the $\displaystyle \frac{1}{128}$ part now and concentrate on the integral itself.

Let us write the integral in the limit form.

$\large \lim_{t\to 0^{+},X\to\infty} \int_{t}^{X} \left(\frac{e^{-u}}{u} - \frac{u}{u^{2}(u^{2}+1)}\right)du$

We know :-

$\displaystyle \int\frac{u}{u^{2}(u^{2}+1)}du = \frac{1}{2}\int\frac{dz}{z(1+z)} = \frac{1}{2}(\ln(z) - \ln(z+1)) = \frac{1}{2}\left(\ln(u^{2}) -\ln(u^{2}+1)\right)$ (without the integration constant)

Applying by parts to the first integral we have:-

$\large \lim_{t\to 0^{+},X\to\infty} \left(e^{-X}\ln(X) - e^{-t}\ln(t) +\int_{t}^{X}e^{-z}\ln(z)dz +\frac{1}{2}\left(\ln(t^{2}) -\ln(t^{2}+1)-\ln(X^{2})+\ln(X^{2}+1)\right)\right)$

$\large = \lim_{t\to 0^{+},X\to\infty} \left(e^{-X}\ln(X) - e^{-t}\ln(t)+\ln(t) +\int_{t}^{X}e^{-z}\ln(z)dz +\frac{1}{2}\left(-\ln(t^{2}+1)-\ln(X^{2})+\ln(X^{2}+1)\right)\right)$

$\large = \lim_{t\to 0^{+},X\to\infty} \left(e^{-X}\ln(X) + \left(1-e^{-t}\right)\ln(t) +\int_{t}^{X}e^{-z}\ln(z)dz +\frac{1}{2}\left(-\ln(t^{2}+1)+\ln(\frac{X^{2}+1}{X^{2}})\right)\right)$

Now let us evaluate the limits one by one :-

Limit 1:- $\displaystyle \lim_{X\to\infty} e^{-X}\ln(X) = 0$

Limit 2:- $\displaystyle \lim_{t\to 0^{+}} (1-e^{-t})\ln(t) =\lim_{t\to 0^{+}} t\ln(t)\cdot \frac{1-e^{-t}}{t} = 0$
Because $\displaystyle \lim_{t \to 0^{+}} t\ln(t) = 0$ and $\displaystyle \lim_{t\to 0^{+}} \frac{1-e^{-t}}{t} = 1$

Limit 3:- $\displaystyle \lim_{t\to 0^{+}} \ln(t^{2}+1) = \ln(1) = 0$

Limit 4:- $\displaystyle \lim_{X\to \infty} \ln(\frac{X^{2}+1}{X^{2}}) = \ln(1) = 0$

I'll leave it upto the readers to prove the limits 1 to 4 . They are pretty elementary and should be easy to prove using taylor series or L'Hopital .

So we have only one limit left and it is

$\large \lim_{t\to 0^{+} , X\to\infty} \int_{t}^{X} e^{-z}\ln(z) dz = \Gamma'(1) = \Gamma(1)\psi(1) = \psi(1) = -\gamma = -0.57721$

Here $\Gamma(x)$ denotes the Gamma Function and $\Gamma'(x)$ denotes it's derivative wrt $x$ and $\psi(x)$ denotes the Digamma Function .

The integral $\displaystyle \int_{0}^{\infty}e^{-z}\ln(z)dz$ is a well known integral . Here is a great youtube video by Dr.Peyam evaluating this integral. Dr.Peyam integral video .

So our answer is :- $\displaystyle \frac{-\gamma}{128} = -0.0045$