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Question Number 122448 by mohammad17 last updated on 17/Nov/20

$$\int \frac{e^x}{e^{{(x+2)}^2}-1}dx$$

Commented by mohammad17 last updated on 17/Nov/20

$$helpmesir$$

Commented by MJS_new last updated on 17/Nov/20

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{this}$$

Commented by Dwaipayan Shikari last updated on 17/Nov/20

$$\frac{1}{e^2}\int \frac{e^t}{e^{t^2}-1}dtt=x+2$$$$\frac{1}{e^2}\int e^t\underset{n=1}{\sum ^{\mathrm{\infty }}}{e}^{-{nt}^2}dt$$$$\frac{1}{e^2}\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\int e^{-{nt}^2+t}dt=\frac{1}{e^2}\underset{n⩾1}{\sum ^{\mathrm{\infty }}}\int e^{-n({(t-\frac{1}{2n})}^2-\frac{1}{4n^2})}dt$$$$=\frac{1}{e^2}\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{e}^{\frac{1}{4n}}\int e^{-n{(t-\frac{1}{2n})}^2}dt(t-\frac{1}{2n})=u\Rightarrow 1=\frac{du}{dt}$$$$=\frac{1}{e^2}\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{e}^{\frac{1}{4n}}\int e^{-{nu}^2}du{nu}^2=p\Rightarrow 2nu=\frac{dp}{du}$$$$=\frac{1}{e^2}\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{e}^{\frac{1}{4n}}\int \underset{k⩾0}{\sum ^{\mathrm{\infty }}}\frac{{(-{nu}^2)}^k}{k!}du$$$$=\frac{1}{e^2}\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{e}^{\frac{1}{4n}}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}\frac{{(-n)}^k}{k!}.\frac{u^{2k+1}}{2k+1}=\frac{1}{e^2}\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{e}^{\frac{1}{4n}}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}\frac{{(-n)}^k}{k!}.\frac{1}{2k+1}{(x-\frac{1}{2n}-2)}^{2k+1}$$$$=\frac{1}{e^2}\underset{n⩾1}{\sum ^{\mathrm{\infty }}}{e}^{\frac{1}{4n}}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}\frac{{(-n)}^k}{k!}.\frac{{(x-\frac{1}{2n}-2)}^{2k+1}}{2k+1}+\frac{C}{e^2}\underset{n⩾0}{\sum ^{\mathrm{\infty }}}{e}^{\frac{1}{4n}}\underset{k⩾0}{\sum ^{\mathrm{\infty }}}\frac{{(-n)}^k}{k!}$$

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