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PhoenixFire
 one year ago
\[\int_{e^1}^{e} {\frac{1}{t(1+\left  ln(t) \right )} dt}\]
Can anyone guide me on starting this integral? I've tried numerous times but keep getting a division by zero.
No need to answer the full question, just a hint on starting would be helpful.
PhoenixFire
 one year ago
\[\int_{e^1}^{e} {\frac{1}{t(1+\left  ln(t) \right )} dt}\] Can anyone guide me on starting this integral? I've tried numerous times but keep getting a division by zero. No need to answer the full question, just a hint on starting would be helpful.

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divu.mkr
 one year ago
Best ResponseYou've already chosen the best response.0tell me what have you done untill now?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0substitute 1+ln t = x hace u donee this ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0and if u have done that, have you changed the limits ?

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0I have tried x=1+ln(t), the limits become 2 and 0. when you integrate 1/x you get lnx.. but you can't put in the limit of 0 into that. this is what I kept getting stuck on.

divu.mkr
 one year ago
Best ResponseYou've already chosen the best response.0then take the limits of 0... that what we do..

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0\(\lim \limits_{y \rightarrow 0} \int \limits_y^2dx/x\)

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I have never been taught such things. I'll give that a go.

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0For \(ln(t)\gt 0\) \[\lim \limits_{y \rightarrow 0} \int \limits_y^2dx/x\]\[=ln(2)\lim \limits_{y \rightarrow 0}ln(y)\] For \(ln(t)\lt 0\) \[\lim \limits_{y \rightarrow 0} \int \limits_y^2dx/x\]\[=\lim \limits_{y \rightarrow 0}ln(y)ln(2)\] The limit of \(ln(y)\) as \(y\rightarrow 0\) is \(\infty\). So how does this work? I have two solutions both at infinity... or have I messed up somewhere?
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