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\[\LARGE \int_{-1}^{0}~\frac{e^{1/x}}{x^3}~dx\]
sub u = 1/x
try letting u=1/x so x=1/u

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Other answers:

then do parts
Guys guys... In the eyes of god, mathematics doesn't exist! Now, let's all hold hands and pray.
yeah u can't avoid parts i think
mathematics is a language and God gave us languages
so shut your 3.14 hole
LOLOL ^
:)
\[\LARGE \int_{-1}^{0}~\frac{e^{1/x}}{x^3}~dx\] \[\LARGE =\int_{-1}^{0}~\frac{1}{x}~e^{1/x}~\frac{1}{x^2}~dx\] \(\Large u=1/x\) and \(\Large du=-\frac{1}{x^2}dx\) \[\LARGE \int_{-1}^{0}~ue^udu\] And then use parts like dan said?
*\[\LARGE -\int_{-1}^{0}~ue^udu\]
you should have an improper integral
\[\LARGE \LARGE \lim_{t \rightarrow 0} \int_{-1}^{t}~ue^udu~\]?
of x is between -1 and 0 and u=1/x |dw:1402252341171:dw|
as we approach 0 from the left what is u getting approaching?
|dw:1402252495152:dw|
we are looking to the left of 0 because our interval is -1 to 0 none of our values occur to the right of 0 so we won't bother to look there
i'm asking him to evaluate the following: \[\lim_{x \rightarrow 0^-}\frac{1}{x}\]
lugi0210 just read the graph above as our x values get near the 0 number (from the left) what do your u values (or if you want call them y values) do?
\(\Large -\infty\)?
right so i'm going to change one thing in your lim from earlier \[\LARGE \LARGE \lim_{t \rightarrow - \infty} \int\limits_{-1}^{t}~ue^udu~ \]
i think there should be a negative in front of that
\[- \LARGE \LARGE \lim_{t \rightarrow - \infty} \int\limits\limits_{-1}^{t}~ue^udu~\]
you know from the du=-1/x^2 dx thing
now use integration by parts then evaluate the limit
Alright, I see it now, it was the limits that was throwing me off >_< Thank you myininaya :)
if u feel lazy about changing the bounds, you may try below : 1) Find the indefinite integral first 2) plugin the bounds
@Luigi0210 did you get it?

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