Question-121397

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} d x =

[by parts]

= 2^{1 / x} x + \ln 2 \int \frac{2^{1 / x}}{x} d x =

[t = \frac{1}{x} \to d x = - x^{2} d t]

= 2^{1 / x} x - \ln 2 \int \frac{2^{t}}{t} d t =

= 2^{1 / x} x - \ln 2 Ei (\ln 2 t) =

= 2^{1 / x} x - \ln 2 Ei (\frac{\ln 2}{x}) + C

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

at form of serie

\int 2^{\frac{1}{x}} dx = \int e^{\frac{1}{x} \ln (2)} = \int \sum_{n = 0}^{\infty} \frac{{(\frac{\ln 2}{x})}^{n}}{n!} = \int \sum_{n = 0}^{\infty} \frac{{(\ln 2)}^{n}}{n! x^{n}}

= \sum_{n = 0}^{\infty} \frac{{(\ln 2)}^{n}}{n!} \int x^{- n} dx = \sum_{n = 0}^{\infty} \frac{{(\ln 2)}^{n}}{n!} \frac{x^{1 - n}}{1 - n} + C

= \sum_{n = 0}^{\infty} \frac{{(\ln 2)}^{n}}{(1 - n) (n!) x^{n - 1}} + C