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Question Number 84879 by sahnaz last updated on 17/Mar/20

e^(∫((2dx)/(xlnx)))

e2dxxlnx

Commented by jagoll last updated on 17/Mar/20

∫ ((2dx)/(x lnx)) = ∫ ((2d(lnx))/(lnx)) = ∫ 2(du/u)  = 2 ln u + c , [u = ln x ]  = 2ln(lnx) + 2lnC = 2ln(Clnx)  e^(∫ ((2dx)/(lnx)))  = e^([ln(Clnx )^2  ])  = (Cln x)^2

2dxxlnx=2d(lnx)lnx=2duu=2lnu+c,[u=lnx]=2ln(lnx)+2lnC=2ln(Clnx)e2dxlnx=e[ln(Clnx)2]=(Clnx)2

Answered by bshahid010@gmail.com last updated on 17/Mar/20

∫_e ^2  (dx/(x.lnx)) put lnx=t⇒(1/x)dx=dt  ∫_e ^2 (dt/t)⇒[lnt]_e ^2 =[ln(lnx)]_e ^2 =ln(ln(2))−ln(ln(e))  ln(ln(2))

2edxx.lnxputlnx=t1xdx=dtMissing \left or extra \rightln(ln(2))

Commented by john santu last updated on 17/Mar/20

the question e^(∫ ((2dx)/(x ln x)))  sir

thequestione2dxxlnxsir

Answered by bshahid010@gmail.com last updated on 18/Mar/20

e^(∫((2dx)/(xlnx )))  put lnx=t⇒(1/x)dx=dt  e^(2∫(dt/(t   )))   =e^(2ln(t)+c)       e^(2ln(lnx)+c)

e2dxxlnxputlnx=t1xdx=dte2dtt=e2ln(t)+ce2ln(lnx)+c

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