Question Number 200984 by sonukgindia last updated on 27/Nov/23
Answered by Calculusboy last updated on 29/Nov/23
$$\mathrm{2} \\ $$
Answered by witcher3 last updated on 29/Nov/23
$$\mathrm{W}\left(\mathrm{x}\right)=\mathrm{y} \\ $$$$\mathrm{x}=\mathrm{ye}^{\mathrm{y}} \Rightarrow\mathrm{dx}=\left(\mathrm{y}+\mathrm{1}\right)\mathrm{e}^{\mathrm{y}} \\ $$$$\mathrm{I}=\int_{\mathrm{W}\left(\mathrm{ln}\left(\frac{\varphi}{\varphi^{\varphi} }\right)\right)} ^{\mathrm{W}\left(\mathrm{2}\left(\mathrm{1}+\varphi\right)\mathrm{ln}\left(\varphi\right)\right)} \mathrm{e}^{\mathrm{y}} \mathrm{dy}\:\: \\ $$$$\frac{\mathrm{2}\left(\mathrm{1}+\varphi\right)\mathrm{ln}\left(\varphi\right)}{\mathrm{W}\left(\left(\varphi^{\mathrm{2}} \right)\mathrm{ln}\left(\varphi^{\mathrm{2}} \right)\right)}−\frac{\mathrm{ln}\left(\frac{\varphi}{\varphi^{\varphi} }\right)}{\mathrm{W}\left(\mathrm{ln}\left(\frac{\varphi}{\varphi^{\varphi} }\right)\right.} \\ $$$$\mathrm{W}\left(\mathrm{ln}\left(\varphi^{\mathrm{1}−\varphi} \right)\right),\mathrm{1}−\varphi=−\frac{\mathrm{1}}{\varphi} \\ $$$$\mathrm{ln}\left(\varphi^{\mathrm{1}−\varphi} \right)=\frac{\mathrm{1}}{\varphi}\mathrm{ln}\left(\frac{\mathrm{1}}{\varphi}\right),\mathrm{W}\left(\mathrm{ln}\varphi^{\mathrm{1}−\varphi} \right)=\frac{\mathrm{1}}{\varphi} \\ $$$$\mathrm{I}=\mathrm{1}+\varphi−\frac{\left(\varphi−\mathrm{1}\right)\mathrm{ln}\left(\frac{\mathrm{1}}{\varphi}\right)}{\mathrm{ln}\left(\frac{\mathrm{1}}{\varphi}\right)}=\mathrm{2} \\ $$$$ \\ $$$$ \\ $$
Answered by Calculusboy last updated on 02/Dec/23
![Solution: let firstly simplify the upper limits and lower limits In(𝛟^2 )+2In(𝛟^𝛟 )=2In𝛟+2𝛟In𝛟=2In𝛟(1+𝛟) In𝛟−In(𝛟^𝛟 )=In𝛟−𝛟In𝛟=In𝛟(1−𝛟) then let u=W(x) x=W(x)∙e^(W(x)) x=u∙e^u dx=(ue^u +e^u )du ⇔ dx=e^u (1+u)du ⇒ ∫_(W(x)) ^(W(x)) (1/((1+u)))e^u ∙(1+u)du I=∫_(W(x)) ^(W(x)) e^u du=e^u ∣_(W(x)) ^(W(x)) +C I=e^(W[2In𝛟(1+𝛟)]) −e^(W[In𝛟(1−𝛟)]) I=𝛟+1−(𝛟−1) I=2](https://www.tinkutara.com/question/Q201220.png)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{{firstly}}\:\boldsymbol{{simplify}}\:\boldsymbol{{the}}\:\boldsymbol{{upper}}\:\boldsymbol{{limits}} \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{lower}}\:\boldsymbol{{limits}} \\ $$$$\boldsymbol{{In}}\left(\boldsymbol{\varphi}^{\mathrm{2}} \right)+\mathrm{2}\boldsymbol{{In}}\left(\boldsymbol{\varphi}^{\boldsymbol{\varphi}} \right)=\mathrm{2}\boldsymbol{{In}\varphi}+\mathrm{2}\boldsymbol{\varphi{In}\varphi}=\mathrm{2}\boldsymbol{{In}\varphi}\left(\mathrm{1}+\boldsymbol{\varphi}\right) \\ $$$$\boldsymbol{{In}\varphi}−\boldsymbol{{In}}\left(\boldsymbol{\varphi}^{\boldsymbol{\varphi}} \right)=\boldsymbol{{In}\varphi}−\boldsymbol{\varphi{In}\varphi}=\boldsymbol{{In}\varphi}\left(\mathrm{1}−\boldsymbol{\varphi}\right) \\ $$$$\boldsymbol{{then}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{W}}\left(\boldsymbol{{x}}\right)\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}=\boldsymbol{{W}}\left(\boldsymbol{{x}}\right)\centerdot\boldsymbol{{e}}^{\boldsymbol{{W}}\left(\boldsymbol{{x}}\right)} \\ $$$$\boldsymbol{{x}}=\boldsymbol{{u}}\centerdot\boldsymbol{{e}}^{\boldsymbol{{u}}} \:\:\:\:\:\:\boldsymbol{{dx}}=\left(\boldsymbol{{ue}}^{\boldsymbol{{u}}} +\boldsymbol{{e}}^{\boldsymbol{{u}}} \right)\boldsymbol{{du}}\:\:\Leftrightarrow\:\:\boldsymbol{{dx}}=\boldsymbol{{e}}^{\boldsymbol{{u}}} \left(\mathrm{1}+\boldsymbol{{u}}\right)\boldsymbol{{du}} \\ $$$$\Rightarrow\:\:\int_{\boldsymbol{{W}}\left(\boldsymbol{{x}}\right)} ^{\boldsymbol{{W}}\left(\boldsymbol{{x}}\right)} \frac{\mathrm{1}}{\left(\mathrm{1}+\boldsymbol{{u}}\right)}\boldsymbol{{e}}^{\boldsymbol{{u}}} \centerdot\left(\mathrm{1}+\boldsymbol{{u}}\right)\boldsymbol{{du}} \\ $$$$\boldsymbol{{I}}=\int_{\boldsymbol{{W}}\left(\boldsymbol{{x}}\right)} ^{\boldsymbol{{W}}\left(\boldsymbol{{x}}\right)} \boldsymbol{{e}}^{\boldsymbol{{u}}} \boldsymbol{{du}}=\boldsymbol{{e}}^{\boldsymbol{{u}}} \mid_{\boldsymbol{{W}}\left(\boldsymbol{{x}}\right)} ^{\boldsymbol{{W}}\left(\boldsymbol{{x}}\right)} +\boldsymbol{{C}} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{e}}^{\boldsymbol{{W}}\left[\mathrm{2}\boldsymbol{{In}\varphi}\left(\mathrm{1}+\boldsymbol{\varphi}\right)\right]} −\boldsymbol{{e}}^{\boldsymbol{{W}}\left[\boldsymbol{{In}\varphi}\left(\mathrm{1}−\boldsymbol{\varphi}\right)\right]} \\ $$$$\boldsymbol{{I}}=\boldsymbol{\varphi}+\mathrm{1}−\left(\boldsymbol{\varphi}−\mathrm{1}\right) \\ $$$$\boldsymbol{{I}}=\mathrm{2} \\ $$