Question Number 12500 by FilupS last updated on 24/Apr/17
$$\mathrm{Please}\:\mathrm{help}\:\mathrm{explain}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\int{e}^{\frac{\mathrm{1}}{{x}}} {dx} \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 24/Apr/17
![e^(1/x) =t⇒(1/x)=lnt⇒(lnx+c)^′ =lnt⇒ ∫(lnx+c)^′ dx=∫lntdt=tlnt−∫t×(1/t)dt= =(lnt−1)t+C(=lnk) =t.ln((t/e))+lnk=ln[k.((t/e))^t ] ⇒lnx+c(=lnk^′ )=ln(k((t/e))^t )⇒k^′ x=k((t/e))^t x=b.((e^(1/x) /e))^e^(1/x) =b.(e^((1/x)−1) )^e^(1/x) =b.(e^((1−x)/x) )^e^(1/x) m=(1/b)=constant. b=(k/k^′ )>0 lnx=lnb+e^(1/x) .ln(e^((1−x)/x) )=lnb+e^(1/x) .(((1−x)/x)) ln(x/b)=e^(1/x) .(((1−x)/x))⇒e^(1/x) =((xln(mx))/(1−x)) I=∫((xln(mx))/(1−x))dx=∫(((ln(mx))/(1−x))−ln(mx))dx= I_1 =∫ln(mx)dx=xln(mx)−∫x×(m/(mx))dx= =xln(mx)−x+n=x(ln(mx)−1)+n. I_2 =∫((ln(mx))/(1−x))dx=ln(1−x).ln(mx)−∫ln(1−x)×(1/x)dx= I_3 =∫ln(1−x)×(1/x)dx=lnx.ln(1−x)−∫((−lnx)/(1−x))dx= =lnx.ln(1−x)+lnx.ln(1−x)−∫((ln(1−x))/x)dx(=I_3 ) I_3 =lnx.ln(1−x)+p I=I_1 +I_2 +I_3 =x(ln(mx)−1)+n+ +ln(1−x).ln(mx)−I_3 +I_3 ⇒ I=[x+ln(1−x)].ln(mx)−x+n . m,n,p∈R, are constants.](https://www.tinkutara.com/question/Q12505.png)
$${e}^{\frac{\mathrm{1}}{{x}}} ={t}\Rightarrow\frac{\mathrm{1}}{{x}}={lnt}\Rightarrow\left({lnx}+{c}\right)^{'} ={lnt}\Rightarrow \\ $$$$\int\left({lnx}+{c}\right)^{'} {dx}=\int{lntdt}={tlnt}−\int{t}×\frac{\mathrm{1}}{{t}}{dt}= \\ $$$$=\left({lnt}−\mathrm{1}\right){t}+\boldsymbol{{C}}\left(={lnk}\right) \\ $$$$={t}.{ln}\left(\frac{{t}}{{e}}\right)+{lnk}={ln}\left[{k}.\left(\frac{{t}}{{e}}\right)^{{t}} \right] \\ $$$$\Rightarrow{lnx}+{c}\left(={lnk}^{'} \right)={ln}\left({k}\left(\frac{{t}}{{e}}\right)^{{t}} \right)\Rightarrow\boldsymbol{{k}}^{'} \boldsymbol{{x}}=\boldsymbol{{k}}\left(\frac{{t}}{{e}}\right)^{{t}} \\ $$$$\boldsymbol{{x}}={b}.\left(\frac{{e}^{\frac{\mathrm{1}}{{x}}} }{{e}}\right)^{\boldsymbol{{e}}^{\frac{\mathrm{1}}{\boldsymbol{{x}}}} } ={b}.\left({e}^{\frac{\mathrm{1}}{{x}}−\mathrm{1}} \right)^{{e}^{\frac{\mathrm{1}}{{x}}} } ={b}.\left({e}^{\frac{\mathrm{1}−{x}}{{x}}} \:\:\right)^{{e}^{\frac{\mathrm{1}}{{x}}} } \\ $$$${m}=\frac{\mathrm{1}}{{b}}={constant}.\:\:\:\:{b}=\frac{{k}}{{k}^{'} }>\mathrm{0} \\ $$$${lnx}={lnb}+{e}^{\frac{\mathrm{1}}{{x}}} .{ln}\left({e}^{\frac{\mathrm{1}−{x}}{{x}}} \right)={lnb}+{e}^{\frac{\mathrm{1}}{{x}}} .\left(\frac{\mathrm{1}−{x}}{{x}}\right) \\ $$$${ln}\frac{{x}}{{b}}={e}^{\frac{\mathrm{1}}{{x}}} .\left(\frac{\mathrm{1}−{x}}{{x}}\right)\Rightarrow{e}^{\frac{\mathrm{1}}{{x}}} =\frac{{xln}\left({mx}\right)}{\mathrm{1}−{x}} \\ $$$$\boldsymbol{{I}}=\int\frac{{xln}\left({mx}\right)}{\mathrm{1}−{x}}{dx}=\int\left(\frac{{ln}\left({mx}\right)}{\mathrm{1}−{x}}−{ln}\left({mx}\right)\right){dx}= \\ $$$${I}_{\mathrm{1}} =\int{ln}\left({mx}\right){dx}={xln}\left({mx}\right)−\int{x}×\frac{{m}}{{mx}}{dx}= \\ $$$$={xln}\left({mx}\right)−{x}+\boldsymbol{{n}}=\boldsymbol{{x}}\left(\boldsymbol{{ln}}\left(\boldsymbol{{mx}}\right)−\mathrm{1}\right)+\boldsymbol{{n}}. \\ $$$$\boldsymbol{{I}}_{\mathrm{2}} =\int\frac{{ln}\left({mx}\right)}{\mathrm{1}−{x}}{dx}={ln}\left(\mathrm{1}−{x}\right).{ln}\left({mx}\right)−\int{ln}\left(\mathrm{1}−{x}\right)×\frac{\mathrm{1}}{{x}}{dx}= \\ $$$${I}_{\mathrm{3}} =\int{ln}\left(\mathrm{1}−{x}\right)×\frac{\mathrm{1}}{{x}}{dx}={lnx}.{ln}\left(\mathrm{1}−{x}\right)−\int\frac{−{lnx}}{\mathrm{1}−{x}}{dx}= \\ $$$$={lnx}.{ln}\left(\mathrm{1}−{x}\right)+{lnx}.{ln}\left(\mathrm{1}−{x}\right)−\int\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}\left(={I}_{\mathrm{3}} \right) \\ $$$${I}_{\mathrm{3}} ={lnx}.{ln}\left(\mathrm{1}−{x}\right)+{p} \\ $$$$\boldsymbol{{I}}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} +{I}_{\mathrm{3}} =\boldsymbol{{x}}\left(\boldsymbol{{ln}}\left(\boldsymbol{{mx}}\right)−\mathrm{1}\right)+\boldsymbol{{n}}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{x}}\right).\boldsymbol{{ln}}\left(\boldsymbol{{mx}}\right)−\boldsymbol{{I}}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\boldsymbol{{I}}_{\mathrm{3}} \Rightarrow \\ $$$$\boldsymbol{{I}}=\left[\boldsymbol{{x}}+\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{x}}\right)\right].\boldsymbol{{ln}}\left(\boldsymbol{{mx}}\right)−\boldsymbol{{x}}+\boldsymbol{{n}}\:. \\ $$$$\boldsymbol{{m}},\boldsymbol{{n}},\boldsymbol{{p}}\in{R},\:\boldsymbol{{are}}\:\boldsymbol{{constants}}. \\ $$
Commented by mrW1 last updated on 24/Apr/17
$${Can}\:{you}\:{check}\:{if}\:\frac{{dI}}{{dx}}={e}^{\frac{\mathrm{1}}{{x}}} \:? \\ $$
Commented by ajfour last updated on 24/Apr/17
$${line}\:#\mathrm{3}\::\:\:\left({t}−\mathrm{1}\right)\mathrm{ln}\:{t}\:\:{should}\:{be} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{ln}\:{t}−\mathrm{1}\right){t}\:. \\ $$
Commented by FilupS last updated on 24/Apr/17
$$\mathrm{according}\:\mathrm{to}\:\mathrm{wolframalpha}: \\ $$$$\int{e}^{\frac{\mathrm{1}}{{x}}} {dx}={e}^{\frac{\mathrm{1}}{{x}}} {x}−\mathrm{Ei}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$\mathrm{Ei}\left({x}\right)=−\int_{−{x}} ^{\:\infty} \frac{\mathrm{1}}{{t}}{e}^{−{t}} {dt} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 24/Apr/17
$${you}\:{are}\:{right}.{it}\:{is}\:{fixed}. \\ $$