Please-help-explain-how-to-solve-e-1-x-dx-

Question Number 12500 by FilupS last updated on 24/Apr/17

Please help explain how to solve

\int e^{\frac{1}{x}} d x

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 24/Apr/17

e^{\frac{1}{x}} = t \Rightarrow \frac{1}{x} = l n t \Rightarrow {(l n x + c)}^{^{'}} = l n t \Rightarrow

\int {(l n x + c)}^{^{'}} d x = \int l n t d t = t l n t - \int t \times \frac{1}{t} d t =

= (l n t - 1) t + C (= l n k)

= t . l n (\frac{t}{e}) + l n k = l n [k . {(\frac{t}{e})}^{t}]

\Rightarrow l n x + c (= {l n k}^{^{'}}) = l n (k {(\frac{t}{e})}^{t}) \Rightarrow k^{^{'}} x = k {(\frac{t}{e})}^{t}

x = b . {(\frac{e^{\frac{1}{x}}}{e})}^{e^{\frac{1}{x}}} = b . {(e^{\frac{1}{x} - 1})}^{e^{\frac{1}{x}}} = b . {(e^{\frac{1 - x}{x}})}^{e^{\frac{1}{x}}}

m = \frac{1}{b} = c o n s t a n t . b = \frac{k}{k^{^{'}}} > 0

l n x = l n b + e^{\frac{1}{x}} . l n (e^{\frac{1 - x}{x}}) = l n b + e^{\frac{1}{x}} . (\frac{1 - x}{x})

l n \frac{x}{b} = e^{\frac{1}{x}} . (\frac{1 - x}{x}) \Rightarrow e^{\frac{1}{x}} = \frac{x l n (m x)}{1 - x}

I = \int \frac{x l n (m x)}{1 - x} d x = \int (\frac{l n (m x)}{1 - x} - l n (m x)) d x =

I_{1} = \int l n (m x) d x = x l n (m x) - \int x \times \frac{m}{m x} d x =

= x l n (m x) - x + n = x (l n (m x) - 1) + n .

I_{2} = \int \frac{l n (m x)}{1 - x} d x = l n (1 - x) . l n (m x) - \int l n (1 - x) \times \frac{1}{x} d x =

I_{3} = \int l n (1 - x) \times \frac{1}{x} d x = l n x . l n (1 - x) - \int \frac{- l n x}{1 - x} d x =

= l n x . l n (1 - x) + l n x . l n (1 - x) - \int \frac{l n (1 - x)}{x} d x (= I_{3})

I_{3} = l n x . l n (1 - x) + p

I = I_{1} + I_{2} + I_{3} = x (l n (m x) - 1) + n +

+ l n (1 - x) . l n (m x) - I_{3}

+ I_{3} \Rightarrow

I = [x + l n (1 - x)] . l n (m x) - x + n .

m, n, p \in R, a r e c o n s t a n t s .

Commented by mrW1 last updated on 24/Apr/17

C a n y o u c h e c k i f \frac{d I}{d x} = e^{\frac{1}{x}} ?

Commented by ajfour last updated on 24/Apr/17

You can't use 'macro parameter character #' in math mode

(\ln t - 1) t .

Commented by FilupS last updated on 24/Apr/17

according to wolframalpha :

\int e^{\frac{1}{x}} d x = e^{\frac{1}{x}} x - Ei (\frac{1}{x})

Ei (x) = - \int_{- x}^{\infty} \frac{1}{t} e^{- t} d t

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 24/Apr/17

y o u a r e r i g h t . i t i s f i x e d .