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Question Number 154421 by peter frank last updated on 18/Sep/21

$$\int [{\left(\frac{\mathrm{x}}{\mathrm{e}}\right)}^{\mathrm{x}}+{\left(\frac{\mathrm{e}}{\mathrm{x}}\right)}^{\mathrm{x}}]\ln \mathrm{xdx}$$

Answered by puissant last updated on 18/Sep/21

$$Q=\int [{\left(\frac{x}{e}\right)}^x+{\left(\frac{e}{x}\right)}^x]lnxdx$$$$=\int \frac{x^{x}lnx}{e^x}dx+\int \frac{e^{x}lnx}{x^x}dx$$$$I=\int \frac{x^{x}lnx}{e^x}dx=\int e^{xlnx}{e}^{-x}lnxdx$$$$=\int e^{(xlnx-x)}lnxdx$$$$u=xlnx-x\Rightarrow du=lnxdx$$$$\Rightarrow I=\int e^{u}du=e^u+C=e^{x(lnx-1)}+C..$$$$J=\int \frac{e^{x}lnx}{x^x}dx=\int e^x{e}^{-xlnx}lnxdx$$$$=\int e^{(x-xlnx)}lnxdx$$$$t=x-xlnx\to dt=-lnxdx$$$$\Rightarrow J=-\int e^{t}dt=-e^t+C=-e^{x(1-lnx)}+C..$$$$Q=I+J..$$$$$$$$\therefore \because Q=e^{x(lnx-1)}-e^{x(1-lnx)}+C...$$

Commented by peter frank last updated on 19/Sep/21

$$\mathrm{thank}\mathrm{you}$$

Answered by peter frank last updated on 19/Sep/21

$${\left(\frac{\mathrm{x}}{\mathrm{e}}\right)}^{\mathrm{x}}=\mathrm{u}{\left(\frac{\mathrm{e}}{\mathrm{x}}\right)}^{\mathrm{x}}=\frac{1}{\mathrm{u}}$$$$\mathrm{x}[\ln \mathrm{x}-\ln \mathrm{e}]=\ln \mathrm{u}$$$$\mathrm{lnxdx}=\frac{1}{\mathrm{u}}\mathrm{du}$$$$\int [\mathrm{u}+\frac{1}{\mathrm{u}}]\ln \mathrm{xdx}=\int \frac{1+{\mathrm{u}}^2}{\mathrm{u}}.\frac{1}{\mathrm{u}}\mathrm{du}$$$$\int (\mathrm{du}+\frac{1}{{\mathrm{u}}^2}\mathrm{du})$$$$=\mathrm{u}-\frac{1}{\mathrm{u}}+\mathrm{C}$$$${\left(\frac{\mathrm{x}}{\mathrm{e}}\right)}^{\mathrm{x}}-{\left(\frac{\mathrm{e}}{\mathrm{x}}\right)}^{\mathrm{x}}+\mathrm{C}$$$$$$

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