Improper Integral (comparison test q)

[coals]

Homework Statement

Use the Comparison Theorem to determine whether the integral is convergent or divergent.
<br /> \int \frac{2+ e^{-x} dx}{x} <br />
from 1 to infinity

Wolframalpha tells me this integral diverges, now i just need to know what to compare it to.

The Attempt at a Solution

So far I've said over the interval [1,infinity) <br /> \frac{2+ e^{-x}}{x} &lt;= \frac{e^{-x}}{x}<br />

When i try to take the limit (t->inf) integral (1,t) of \frac{e^{-x}}{x}<br /> i can't integrate this (tried IBP, don't see any obvious trig subs).

Am i just doing it wrong?

Thanks for your help everyone.

That's suposed to be e^(-x) btw i fail at this.

 

[gabbagabbahey]

So far I've said over the interval [1,infinity) <br /> \frac{2+ e^{-x}}{x} &lt;= \frac{e^{-x}}{x}<br />

You might want to recheck the direction of that \leq sign ... Is e^{-x}[ ever greater than one on your interval? If not, then surely you can say \frac{2+ e^{-x}}{x} \leq \frac{3}{x} [/tex]

 

[Office_Shredder]

To expand on gabba's post:
We know that e^{-x} is integrable, so dividing it by isn't likely to change anything. Since we know (from Wolfram) that the integral diverges, we should look at the other half of the function (the constant divided by x)

 

[Bohrok]

Why not start with the comparison 2 + e^-x > 1, then divide by x and compare the integrals?

 

[System]

(2+e^(-x))/x > 2/x

 

[coals]

Thank you all for your help. Very appreciated , i got the answer and i understand what to do now in similar situations. Doing function comparisons seems to be my weak spot I'm alright with everything else but i suck at these.