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log-x-1-x-x-dx-




Question Number 151501 by talminator2856791 last updated on 21/Aug/21
                        ∫ (log x + 1)x^x  dx
$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\left(\mathrm{log}\:{x}\:+\:\mathrm{1}\right){x}^{{x}} \:{dx} \\ $$$$\: \\ $$
Answered by puissant last updated on 21/Aug/21
K=∫(1+lnx)x^x dx  x^x =e^(xlnx)   (d/dx)e^(xlnx) =(1+lnx)x^x   ⇒ K= x^x +C
$${K}=\int\left(\mathrm{1}+{lnx}\right){x}^{{x}} {dx} \\ $$$${x}^{{x}} ={e}^{{xlnx}} \:\:\frac{{d}}{{dx}}{e}^{{xlnx}} =\left(\mathrm{1}+{lnx}\right){x}^{{x}} \\ $$$$\Rightarrow\:{K}=\:{x}^{{x}} +{C} \\ $$
Commented by Olaf_Thorendsen last updated on 21/Aug/21
x^x  = e^(xlnx)   (d/dx)x^x  = (d/dx)e^(xlnx)  = (1+lnx)e^(xlnx)  = (1+lnx)x^x   K = x^x +C
$${x}^{{x}} \:=\:{e}^{{x}\mathrm{ln}{x}} \\ $$$$\frac{{d}}{{dx}}{x}^{{x}} \:=\:\frac{{d}}{{dx}}{e}^{{x}\mathrm{ln}{x}} \:=\:\left(\mathrm{1}+\mathrm{ln}{x}\right){e}^{{x}\mathrm{ln}{x}} \:=\:\left(\mathrm{1}+\mathrm{ln}{x}\right){x}^{{x}} \\ $$$$\mathrm{K}\:=\:{x}^{{x}} +\mathrm{C} \\ $$
Commented by Tawa11 last updated on 22/Aug/21
Sir, help me check    Q151636
$$\mathrm{Sir},\:\mathrm{help}\:\mathrm{me}\:\mathrm{check}\:\:\:\:\mathrm{Q151636} \\ $$

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