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Question Number 122938 by CanovasCamiseros last updated on 21/Nov/20

Commented by CanovasCamiseros last updated on 21/Nov/20

$$\mathit{h}\mathit{e}\mathit{l}\mathit{p}$$

Answered by Dwaipayan Shikari last updated on 21/Nov/20

$$\int 2^{\frac{1}{x}}dx$$$$=\int e^{\frac{1}{x}log\left(2\right)}dx$$$$=\int \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{x}log2\right)}^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{log}^n\left(2\right)}{n!}.\int {\left(\frac{1}{x}\right)}^{n}dx\frac{1}{x}=t\Rightarrow -\frac{1}{x^2}=\frac{dt}{dx}$$$$=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{log}^n\left(2\right)}{n!}\int t^{n-2}dt$$$$=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{log}^n\left(2\right)}{n!}.\frac{t^{n-1}}{n-1}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{log}^n\left(2\right)}{n!}.\frac{t^{n-1}}{1-n}$$

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