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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
(logx+1)xxdx
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=(1+lnx)xxdxxx=exlnxddxexlnx=(1+lnx)xxK=xx+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
xx=exlnxddxxx=ddxexlnx=(1+lnx)exlnx=(1+lnx)xxK=xx+C
Commented by Tawa11 last updated on 22/Aug/21
Sir, help me check    Q151636
Sir,helpmecheckQ151636

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