## PhoenixFire Group Title $\int_{e^-1}^{e} {\frac{1}{t(1+\left | ln(t) \right |)} dt}$ 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. 11 months ago 11 months ago

1. divu.mkr Group Title

tell me what have you done untill now?

2. hartnn Group Title

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

3. hartnn Group Title

**have done

4. hartnn Group Title

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

5. PhoenixFire Group Title

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.

6. divu.mkr Group Title

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

7. hartnn Group Title

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

8. PhoenixFire Group Title

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

9. PhoenixFire Group Title

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?

10. PhoenixFire Group Title

@hartnn