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

$$\mathrm{Please}\mathrm{help}\mathrm{explain}\mathrm{how}\mathrm{to}\mathrm{solve}$$$$\int e^{\frac{1}{x}}dx$$

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

$$e^{\frac{1}{x}}=t\Rightarrow \frac{1}{x}=lnt\Rightarrow {(lnx+c)}^{{}^{\prime }}=lnt\Rightarrow$$$$\int {(lnx+c)}^{{}^{\prime }}dx=\int lntdt=tlnt-\int t\times \frac{1}{t}dt=$$$$=(lnt-1)t+\mathit{C}(=lnk)$$$$=t.ln\left(\frac{t}{e}\right)+lnk=ln[k.{\left(\frac{t}{e}\right)}^t]$$$$\Rightarrow lnx+c(={lnk}^{{}^{\prime }})=ln\left(k{\left(\frac{t}{e}\right)}^t\right)\Rightarrow {\mathit{k}}^{{}^{\prime }}\mathit{x}=\mathit{k}{\left(\frac{t}{e}\right)}^t$$$$\mathit{x}=b.{\left(\frac{e^{\frac{1}{x}}}{e}\right)}^{{\mathit{e}}^{\frac{1}{\mathit{x}}}}=b.{\left(e^{\frac{1}{x}-1}\right)}^{e^{\frac{1}{x}}}=b.{\left(e^{\frac{1-x}{x}}\right)}^{e^{\frac{1}{x}}}$$$$m=\frac{1}{b}=constant.b=\frac{k}{k^{{}^{\prime }}}>0$$$$lnx=lnb+e^{\frac{1}{x}}.ln\left(e^{\frac{1-x}{x}}\right)=lnb+e^{\frac{1}{x}}.\left(\frac{1-x}{x}\right)$$$$ln\frac{x}{b}=e^{\frac{1}{x}}.\left(\frac{1-x}{x}\right)\Rightarrow e^{\frac{1}{x}}=\frac{xln\left(mx\right)}{1-x}$$$$\mathit{I}=\int \frac{xln\left(mx\right)}{1-x}dx=\int (\frac{ln\left(mx\right)}{1-x}-ln\left(mx\right))dx=$$$$I_1=\int ln\left(mx\right)dx=xln\left(mx\right)-\int x\times \frac{m}{mx}dx=$$$$=xln\left(mx\right)-x+\mathit{n}=\mathit{x}(\mathit{l}\mathit{n}\left(\mathit{m}\mathit{x}\right)-1)+\mathit{n}.$$$${\mathit{I}}_2=\int \frac{ln\left(mx\right)}{1-x}dx=ln(1-x).ln\left(mx\right)-\int ln(1-x)\times \frac{1}{x}dx=$$$$I_3=\int ln(1-x)\times \frac{1}{x}dx=lnx.ln(1-x)-\int \frac{-lnx}{1-x}dx=$$$$=lnx.ln(1-x)+lnx.ln(1-x)-\int \frac{ln(1-x)}{x}dx(=I_3)$$$$I_3=lnx.ln(1-x)+p$$$$\mathit{I}=I_1+I_2+I_3=\mathit{x}(\mathit{l}\mathit{n}\left(\mathit{m}\mathit{x}\right)-1)+\mathit{n}+$$$$+\mathit{l}\mathit{n}(1-\mathit{x}).\mathit{l}\mathit{n}\left(\mathit{m}\mathit{x}\right)-{\mathit{I}}_3$$$$+{\mathit{I}}_3\Rightarrow$$$$\mathit{I}=[\mathit{x}+\mathit{l}\mathit{n}(1-\mathit{x})].\mathit{l}\mathit{n}\left(\mathit{m}\mathit{x}\right)-\mathit{x}+\mathit{n}.$$$$\mathit{m},\mathit{n},\mathit{p}\in R,\mathit{a}\mathit{r}\mathit{e}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathit{s}.$$

Commented by mrW1 last updated on 24/Apr/17

$$Canyoucheckif\frac{dI}{dx}=e^{\frac{1}{x}}?$$

Commented by ajfour last updated on 24/Apr/17

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$(\ln t-1)t.$$

Commented by FilupS last updated on 24/Apr/17

$$\mathrm{according}\mathrm{to}\mathrm{wolframalpha}:$$$$\int e^{\frac{1}{x}}dx=e^{\frac{1}{x}}x-\mathrm{Ei}\left(\frac{1}{x}\right)$$$$\mathrm{Ei}\left(x\right)=-{\int }_{-x}^{\mathrm{\infty }}\frac{1}{t}{e}^{-t}dt$$

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

$$youareright.itisfixed.$$

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