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Question Number 180268 by Noorzai last updated on 09/Nov/22

Answered by Ar Brandon last updated on 09/Nov/22

Q121680

$$\mathrm{Q121680} \\ $$

Answered by LEKOUMA last updated on 10/Nov/22

∫x^x dx=∫e^(xln x) dx=(xln x)′e^(xln x) +c   avce c∈R                     =(ln x+1)e^(xln x)  +c  avec c∈R  ∫x^x dx=ln x×e^(xln x) +e^(xln x)   +c   avec c∈R

$$\int{x}^{{x}} {dx}=\int{e}^{{x}\mathrm{ln}\:{x}} {dx}=\left({x}\mathrm{ln}\:{x}\right)'{e}^{{x}\mathrm{ln}\:{x}} +{c}\:\:\:{avce}\:{c}\in\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{ln}\:{x}+\mathrm{1}\right){e}^{{x}\mathrm{ln}\:{x}} \:+{c}\:\:{avec}\:{c}\in\mathbb{R} \\ $$$$\int{x}^{{x}} {dx}=\mathrm{ln}\:{x}×{e}^{{x}\mathrm{ln}\:{x}} +{e}^{{x}\mathrm{ln}\:{x}} \:\:+{c}\:\:\:{avec}\:{c}\in\mathbb{R} \\ $$$$ \\ $$

Commented by mr W last updated on 10/Nov/22

if y=ln x×e^(xln x) +e^(xln x)   can you get  (dy/dx)=x^x ?  if no, then your answer is wrong.

$${if}\:{y}=\mathrm{ln}\:{x}×{e}^{{x}\mathrm{ln}\:{x}} +{e}^{{x}\mathrm{ln}\:{x}} \\ $$$${can}\:{you}\:{get}\:\:\frac{{dy}}{{dx}}={x}^{{x}} ? \\ $$$${if}\:{no},\:{then}\:{your}\:{answer}\:{is}\:{wrong}. \\ $$

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