Question

\[\int_{e^-1}^{e} {\frac{1}{t(1+\left | ln(t) \right |)} dt}\]

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.

Answer 1

tell me what have you done untill now?

Answer 2

substitute 1+ln t = x
hace u donee this ?

Answer 3

**have
done

Answer 4

and if u have done that, have you changed the limits ?

Answer 5

I 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.

Answer 6

then take the limits of 0... that what we do..

Answer 7

\(\lim \limits_{y \rightarrow 0} \int \limits_y^2dx/x\)

Answer 8

Okay, I have never been taught such things.
I'll give that a go.

Answer 9

For \(ln(t)\gt 0\)
\[\lim \limits_{y \rightarrow 0} \int \limits_y^2dx/x\]\[=ln(2)-\lim \limits_{y \rightarrow 0}ln(y)\]
For \(ln(t)\lt 0\)
\[\lim \limits_{y \rightarrow 0} \int \limits_y^2-dx/x\]\[=\lim \limits_{y \rightarrow 0}ln(y)-ln(2)\]

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?

Answer 10

@hartnn