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Question Number 71850 by SmNayon11 last updated on 21/Oct/19

$$\int \ln ({\mathrm{x}}^{{\mathrm{x}}^{\mathrm{x}}}.{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{x}}})\mathrm{dx}=?$$

Answered by MJS last updated on 21/Oct/19

$$\int \ln \left(x^{x^x}{\mathrm{e}}^{x^x}\right)dx=\int x^x(1+\ln x)dx=$$$$[t=x^x\to dx=\frac{dt}{x^x(1+\ln x)}]$$$$=\int dt=t=x^x+C$$

Answered by mind is power last updated on 21/Oct/19

$$\int ({\mathrm{x}}^{\mathrm{x}}\ln \left(\mathrm{x}\right)+{\mathrm{x}}^{\mathrm{x}})\mathrm{dx}$$$$\mathrm{let}\mathrm{u}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{x}}={\mathrm{e}}^{\mathrm{xln}\left(\mathrm{x}\right)}\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=(\ln \left(\mathrm{x}\right)+1){\mathrm{e}}^{\mathrm{xln}\left(\mathrm{x}\right)}={\mathrm{x}}^{\mathrm{x}}\ln \left(\mathrm{x}\right)+{\mathrm{x}}^{\mathrm{x}}$$$$\int \ln ({\mathrm{x}}^{{\mathrm{x}}^{\mathrm{x}}}.{\mathrm{e}}^{{\mathrm{x}}^{\mathrm{x}}})\mathrm{dx}=\int \frac{\mathrm{du}}{\mathrm{dx}}.\mathrm{dx}=\int \mathrm{du}=\mathrm{u}+\mathrm{c}={\mathrm{x}}^{\mathrm{x}}+\mathrm{c}$$

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