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$$\int \frac{(z^{-3})e^{1/z}}1dz$$

I use tailor series but I cant solve it also I use common solution but I cant solve it

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    $\begingroup$ Is the integral over some path, or do you just want an antiderivative? And why is there a $1$ in the denominator? $\endgroup$ – Arthur Jun 22 '17 at 12:34
  • $\begingroup$ I just want its antiderivative, I cant write this function correctly because dz goes into denominator !! $\endgroup$ – Nima Gorjinezhad Jun 22 '17 at 12:43
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    $\begingroup$ @NimaGorjinezhad You asked a similar question here. It seems you're having trouble using MathJax. Please go through this tutorial. $\endgroup$ – Sahiba Arora Jun 22 '17 at 12:46
  • $\begingroup$ You mean $\int \frac{\exp(\frac{1}{z})}{z^3 dz}$ ? $\endgroup$ – user392395 Jun 22 '17 at 13:14
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$\int z^{-3}e^{1/z}dz=- \int u(z)v'(z) dz$, where $u(z)=1/z$ and $v(z)=e^{1/z}$.

Your turn !

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The Solution is

$$ \int z^{-3}\, e^{1/z} dz=\frac{e^{\frac{1}{z}} (z-1)}{z}. $$

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