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Question Number 121397 by abdelsalamalmukasabe last updated on 07/Nov/20

Answered by MJS_new last updated on 07/Nov/20

$$\int 2^{1/x}dx=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=2^{1/x}x+\ln 2\int \frac{2^{1/x}}{x}dx=$$$$[t=\frac{1}{x}\to dx=-x^{2}dt]$$$$=2^{1/x}x-\ln 2\int \frac{2^t}{t}dt=$$$$=2^{1/x}x-\ln 2\mathrm{Ei}\left(\ln 2t\right)=$$$$=2^{1/x}x-\ln 2\mathrm{Ei}\left(\frac{\ln 2}{x}\right)+C$$

Answered by mathmax by abdo last updated on 07/Nov/20

$$\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}$$$$\int 2^{\frac{1}{\mathrm{x}}}\mathrm{dx}=\int {\mathrm{e}}^{\frac{1}{\mathrm{x}}\ln \left(2\right)}=\int \sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\left(\frac{\ln 2}{\mathrm{x}}\right)}^{\mathrm{n}}}{\mathrm{n}!}=\int \sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\left(\ln 2\right)}^{\mathrm{n}}}{\mathrm{n}!{\mathrm{x}}^{\mathrm{n}}}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\left(\ln 2\right)}^{\mathrm{n}}}{\mathrm{n}!}\int {\mathrm{x}}^{-\mathrm{n}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\left(\ln 2\right)}^{\mathrm{n}}}{\mathrm{n}!}\frac{{\mathrm{x}}^{1-\mathrm{n}}}{1-\mathrm{n}}+\mathrm{C}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\left(\ln 2\right)}^{\mathrm{n}}}{(1-\mathrm{n})(\mathrm{n}!){\mathrm{x}}^{\mathrm{n}-1}}+\mathrm{C}$$$$$$

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