I want to solve a integral which contains a shift version
$$\int^{\infty}_{c}N [(1-e^{-1/t})]^{N-1} \frac{-1}{(t-c)^2}e^{-1/(t-c)}dt$$
This kind of integral has the form of normal integral
$$ \int _{c} ^{\infty} {N [ f(x) ] ^{N-1}} f(x) ^{'} d(x)=f( \infty) ^{N} -f(c) ^{N} $$
However, in this case, the integral has the form of
$$ \int _{c} ^{\infty} {N [ f(x) ] ^{N-1}} f(x-c) ^{'} d(x), $$
where $$ f(x) = (1-e^{-1/t}), $$
$$ f(x-c)^{'} = \frac{-1}{(t-c)^2}e^{-1/(t-c)}$$
I try to replace ($t-c$) by u, then $x=1/u$ and some more substitutions. However, the integral is still complex and difficult to solve.
$$ \int _{0} ^{\infty} N [1-e^{-x/(1+cx)}]^{N-1} e^{x} dx$$
Can I solve this kind of integral by some estimation techniques or near optimal solutions?
Thank you very much!