Question-122938

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

Commented by CanovasCamiseros last updated on 21/Nov/20

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

∫2^(1/x) dx =∫e^((1/x)log(2)) dx =∫Σ_(n=0) ^∞ ((((1/x)log2)^n )/(n!)) =Σ_(n=0) ^∞ ((log^n (2))/(n!)).∫((1/x))^n dx (1/x)=t⇒−(1/x^2 )=(dt/dx) =−Σ_(n=0) ^∞ ((log^n (2))/(n!))∫t^(n−2) dt =−Σ_(n=0) ^∞ ((log^n (2))/(n!)).(t^(n−1) /(n−1))=Σ_(n=0) ^∞ ((log^n (2))/(n!)).(t^(n−1) /(1−n))

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

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