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Question Number 38726 by maxmathsup by imad last updated on 28/Jun/18

$$letf\left(x\right)=\frac{x+1}{2+e^{-2x}}developpfatintegrserie.$$

Commented by math khazana by abdo last updated on 01/Jul/18

$$f\left(x\right)=\frac{1}{2}(x+1)\frac{1}{1+\left(\frac{e^{-2x}}{2}\right)}=\frac{1}{2}(x+1)\sum _{n=0}^{\mathrm{\infty }}\frac{e^{-2nx}}{2^n}$$$$=\frac{1}{2}(x+1)\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2^n}\left(\sum _{p=0}^{\mathrm{\infty }}\frac{{(-2nx)}^p}{p!}\right)$$$$=\frac{1}{2}(x+1)\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2^n}\left(\sum _{p=0}^{\mathrm{\infty }}\frac{{(-2n)}^p}{p!}{x}^p\right)...$$

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