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PhoenixFire
 2 years 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
 2 years 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.

This Question is Closed

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0tell me what have you done untill now?

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

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

PhoenixFire
 2 years 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.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0then take the limits of 0... that what we do..

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

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

PhoenixFire
 2 years 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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