How can one show that $\int_{0}^{\infty}\left({1\over 1+nx^n}-e^{-nx^n}\right)\cdot{\mathrm dx\over x^{1+n}}=1-\gamma?$ Consider

$$\int_{0}^{\infty}\left({1\over 1+nx^n}-e^{-nx^n}\right)\cdot{\mathrm dx\over x^{1+n}}=1-\gamma\tag1$$
  $n\ge1$;(integers)

n seem to be not involved in the closed form(why?)
How does one show that $(1)$ converges to $1-\gamma?$
An attempt:
$$(1+nx^n)^{-1}=1-nx^n+(nx^n)^2-(nx^n)^3+\cdots$$
$(1)$ becomes
$$\int_{0}^{\infty}\left(x^{-n-1}-nx^{-1}+n^2x^{n-1}-n^3x^{2n-1}+\cdots-{e^{-nx^n}\over x^{n+1}}\right)\mathrm dx\tag2$$
$(2)$ divgerges, how  else can we tackle $(1)$?
Or do we have to differentiate m times w.r.t n
 A: If you change variable $n x^n=y$, almost as suggested by tired in his comment, you end with 
$$\int_{0}^{\infty}\left({1\over 1+nx^n}-e^{-nx^n}\right)\cdot{\mathrm dx\over x^{1+n}}=\int_{0}^{\infty}\frac{\frac{1}{y+1}-e^{-y}}{y^2}\,dy$$ The antiderivative is given by $$\int\frac{\frac{1}{y+1}-e^{-y}}{y^2}\,dy=\text{Ei}(-y)+\frac{e^{-y}}{y}-\frac{1}{y}-\log (y)+\log (y+1)=f(y)$$ where appears the exponential integral.
Since $\lim_{y\to \infty } \, f(y)=0$, we are left with $\lim_{y\to 0 } \, f(y)$.
Using asymptotics , $$\text{Ei}(-y)=\gamma+\log (y) -y+\frac{y^2}{4}+O\left(y^3\right)$$ and Taylor series for small values of $y$
$$f(y)=(\gamma -1)+\frac{y}{2}-\frac{5 y^2}{12}+O\left(y^3\right)$$ which then leads to the result.
A: Claude Leibovicy already send a nice answer, while I was still typing.
Nevertheless, I joint my answer. Even if not smart, solving by brute force :

A: $$I= \int_{0}^{\infty}\frac{\frac{1}{x+1}-e^{-x}}{x^2}\,dx$$
Let 
$$I(s)= \int_{0}^{\infty}\frac{x^{s-1}}{x+1}-x^{s-1}e^{-x}\,dx =\Gamma(s)\Gamma(1-s)-\Gamma(s)$$
Taking the limit 
$$\lim_{s \to -1}\Gamma(s)\Gamma(1-s)-\Gamma(s)  $$
Near $-1$ we have 
$$\Gamma(s)\Gamma(1-s)-\Gamma(s) =  -\frac{1}{s+1}+\frac{1}{s+1}-\gamma+1+O((s+1))$$
Hence 
$$I= \int_{0}^{\infty}\frac{\frac{1}{x+1}-e^{-x}}{x^2}\,dx = 1-\gamma$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\int_{0}^{\infty}\pars{{1 \over 1 + nx^{n}} - \expo{-nx^{n}}}
\,{\dd x\over x^{1 + n}}
\,\,\,\stackrel{y\ =\ nx^{n}}{=}\,\,\,
\int_{0}^{\infty}\pars{{1 \over y + 1} - \expo{-y}}\,{\dd y \over y^{2}}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\bracks{%
\int_{\epsilon}^{\infty}{\dd y \over y^{2}} -
\int_{\epsilon}^{\infty}\pars{{1 \over y} - {1 \over y + 1}}\dd y -
\int_{\epsilon}^{\infty}{\expo{-y} \over y^{2}}\,\dd y}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\bracks{%
{1 \over \epsilon} + \ln\pars{\epsilon \over \epsilon + 1} - {\expo{-\epsilon} \over \epsilon} + \int_{\epsilon}^{\infty}{\expo{-y} \over y}\,\dd y}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\bracks{%
{1 \over \epsilon} + \ln\pars{\epsilon \over \epsilon + 1} - {\expo{-\epsilon} \over \epsilon} - \ln\pars{\epsilon}\expo{-\epsilon} + \int_{\epsilon}^{\infty}\ln\pars{y}\expo{-y}\,\dd y}
\\[5mm] = &
\underbrace{\lim_{\epsilon \to 0^{+}}\bracks{%
{1 \over \epsilon} + \ln\pars{\epsilon \over \epsilon + 1} - {\expo{-\epsilon} \over \epsilon} - \ln\pars{\epsilon}\expo{-\epsilon}}}_{\ds{=\ 1}}\  +\
\underbrace{\int_{0}^{\infty}\ln\pars{y}\expo{-y}\,\dd y}_{\ds{=\ -\gamma}} =
\bbx{\ds{1 - \gamma}}
\end{align}
