Question Number 154421 by peter frank last updated on 18/Sep/21
![∫[((x/e))^x +((e/x))^x ]ln xdx](https://www.tinkutara.com/question/Q154421.png)
$$\int\left[\left(\frac{\mathrm{x}}{\mathrm{e}}\right)^{\mathrm{x}} +\left(\frac{\mathrm{e}}{\mathrm{x}}\right)^{\mathrm{x}} \right]\mathrm{ln}\:\mathrm{xdx} \\ $$
Answered by puissant last updated on 18/Sep/21
![Q=∫[((x/e))^x +((e/x))^x ]lnx dx =∫((x^x lnx)/e^x )dx + ∫((e^x lnx)/x^x )dx I=∫((x^x lnx)/e^x )dx = ∫e^(xlnx) e^(−x) lnx dx =∫e^((xlnx−x)) lnx dx u=xlnx−x ⇒ du=lnxdx ⇒ I=∫e^u du = e^u +C = e^(x(lnx−1)) +C.. J=∫((e^x lnx)/x^x )dx = ∫e^x e^(−xlnx) lnx dx =∫e^((x−xlnx)) lnx dx t=x−xlnx → dt=−lnxdx ⇒ J=−∫e^t dt = −e^t +C = −e^(x(1−lnx)) +C.. Q = I + J.. ∴∵ Q= e^(x(lnx−1)) −e^(x(1−lnx)) +C...](https://www.tinkutara.com/question/Q154426.png)
$${Q}=\int\left[\left(\frac{{x}}{{e}}\right)^{{x}} +\left(\frac{{e}}{{x}}\right)^{{x}} \right]{lnx}\:{dx}\: \\ $$$$=\int\frac{{x}^{{x}} {lnx}}{{e}^{{x}} }{dx}\:+\:\int\frac{{e}^{{x}} {lnx}}{{x}^{{x}} }{dx} \\ $$$${I}=\int\frac{{x}^{{x}} {lnx}}{{e}^{{x}} }{dx}\:=\:\int{e}^{{xlnx}} {e}^{−{x}} {lnx}\:{dx} \\ $$$$=\int{e}^{\left({xlnx}−{x}\right)} {lnx}\:{dx}\: \\ $$$${u}={xlnx}−{x}\:\Rightarrow\:{du}={lnxdx} \\ $$$$\Rightarrow\:{I}=\int{e}^{{u}} {du}\:=\:{e}^{{u}} +{C}\:=\:{e}^{{x}\left({lnx}−\mathrm{1}\right)} +{C}.. \\ $$$${J}=\int\frac{{e}^{{x}} {lnx}}{{x}^{{x}} }{dx}\:=\:\int{e}^{{x}} {e}^{−{xlnx}} {lnx}\:{dx} \\ $$$$=\int{e}^{\left({x}−{xlnx}\right)} {lnx}\:{dx}\: \\ $$$${t}={x}−{xlnx}\:\rightarrow\:{dt}=−{lnxdx} \\ $$$$\Rightarrow\:{J}=−\int{e}^{{t}} {dt}\:=\:−{e}^{{t}} +{C}\:=\:−{e}^{{x}\left(\mathrm{1}−{lnx}\right)} +{C}.. \\ $$$$\:\:\:\:\:\:{Q}\:=\:{I}\:+\:{J}.. \\ $$$$ \\ $$$$\therefore\because\:\:{Q}=\:{e}^{{x}\left({lnx}−\mathrm{1}\right)} −{e}^{{x}\left(\mathrm{1}−{lnx}\right)} +{C}… \\ $$
Commented by peter frank last updated on 19/Sep/21
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by peter frank last updated on 19/Sep/21
![((x/e))^x =u ((e/x))^x =(1/u) x[ln x−ln e]=ln u lnxdx=(1/u)du ∫[u+(1/u)]ln xdx=∫((1+u^2 )/u).(1/u)du ∫(du+(1/u^2 )du) =u−(1/u)+C ((x/e))^x −((e/x))^x +C](https://www.tinkutara.com/question/Q154558.png)
$$\left(\frac{\mathrm{x}}{\mathrm{e}}\right)^{\mathrm{x}} =\mathrm{u}\:\:\:\:\:\:\left(\frac{\mathrm{e}}{\mathrm{x}}\right)^{\mathrm{x}} =\frac{\mathrm{1}}{\mathrm{u}} \\ $$$$\mathrm{x}\left[\mathrm{ln}\:\mathrm{x}−\mathrm{ln}\:\mathrm{e}\right]=\mathrm{ln}\:\mathrm{u} \\ $$$$\mathrm{lnxdx}=\frac{\mathrm{1}}{\mathrm{u}}\mathrm{du} \\ $$$$\int\left[\mathrm{u}+\frac{\mathrm{1}}{\mathrm{u}}\right]\mathrm{ln}\:\mathrm{xdx}=\int\frac{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }{\mathrm{u}}.\frac{\mathrm{1}}{\mathrm{u}}\mathrm{du} \\ $$$$\int\left(\mathrm{du}+\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }\mathrm{du}\right) \\ $$$$=\mathrm{u}−\frac{\mathrm{1}}{\mathrm{u}}+\mathrm{C} \\ $$$$\left(\frac{\mathrm{x}}{\mathrm{e}}\right)^{\mathrm{x}} −\left(\frac{\mathrm{e}}{\mathrm{x}}\right)^{\mathrm{x}} +\mathrm{C} \\ $$$$ \\ $$