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

∫ln(x^x^x  .e^x^x  )dx=?

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

Answered by MJS last updated on 21/Oct/19

∫ln (x^x^x  e^x^x  )dx=∫x^x (1+ln x)dx=       [t=x^x  → dx=(dt/(x^x (1+ln x)))]  =∫dt=t=x^x +C

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

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

∫(x^x ln(x)+x^x )dx  let u(x)=x^x =e^(xln(x)) ⇒(du/dx)=(ln(x)+1)e^(xln(x)) =x^x ln(x)+x^x   ∫ln(x^x^x  .e^x^x  )dx=∫(du/dx).dx=∫du=u+c=x^x +c

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

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