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Question-76317




Question Number 76317 by Master last updated on 26/Dec/19
Commented by john santu last updated on 26/Dec/19
∫e^(xlnx)  dx. with integration by part  u=e^(xlnx) →du=(lnx+1)e^(xlnx) . dv=dx   I=xe^(xlnx) −∫(xlnx+x)e^(xlnx) dx
$$\int{e}^{{xlnx}} \:{dx}.\:{with}\:{integration}\:{by}\:{part} \\ $$$${u}={e}^{{xlnx}} \rightarrow{du}=\left({lnx}+\mathrm{1}\right){e}^{{xlnx}} .\:{dv}={dx}\: \\ $$$${I}={xe}^{{xlnx}} −\int\left({xlnx}+{x}\right){e}^{{xlnx}} {dx} \\ $$
Commented by turbo msup by abdo last updated on 26/Dec/19
at form of serie  ∫ x^x dx =∫ e^(xln(x)) dx  =∫  Σ_(n=0) ^∞   ((x^n (lnx)^n )/(n!)) dx  =Σ_(n=0) ^∞  (1/(n!)) ∫ x^n (lnx)^n  dx  and a_n =∫x^n (lnx)^n  dx can be  fiund by recurrence...
$${at}\:{form}\:{of}\:{serie} \\ $$$$\int\:{x}^{{x}} {dx}\:=\int\:{e}^{{xln}\left({x}\right)} {dx} \\ $$$$=\int\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} \left({lnx}\right)^{{n}} }{{n}!}\:{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:\int\:{x}^{{n}} \left({lnx}\right)^{{n}} \:{dx} \\ $$$${and}\:{a}_{{n}} =\int{x}^{{n}} \left({lnx}\right)^{{n}} \:{dx}\:{can}\:{be} \\ $$$${fiund}\:{by}\:{recurrence}… \\ $$
Commented by mr W last updated on 27/Dec/19
i see absolutely no sense for   repeated posting questions like  ∫x^x dx or ∫x!dx.
$${i}\:{see}\:{absolutely}\:{no}\:{sense}\:{for}\: \\ $$$${repeated}\:{posting}\:{questions}\:{like} \\ $$$$\int{x}^{{x}} {dx}\:{or}\:\int{x}!{dx}. \\ $$

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