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Integral of (e^x)/x

Mathematics
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Applying the Divide rule of Derivative here..
lol water do it by that way i will give you thousand medals :P
and its integration man~

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Other answers:

Sorry I did not read the question fully..
\[\large \int\limits\frac{e^{x}}{x} dx\]?? looks like cant be solved by elementary methods..
ahaha omni in your dreams do it :P
We can use here Integration by Parts...
http://www.wolframalpha.com/input/?i=integrate+%28e^x%29%2Fx like srsly? how can you solve it?
hahahah omni i heard you saying i own calculus ? :P
term by term integration of the infinite series?
like srsly.. http://en.wikipedia.org/wiki/Exponential_integral
hi water , have you heard of exp x = \[x + x ^{2}/2! + x ^{3}/3! + x ^{4}/4!.........\]
Wasi, what calculus are you in
if you know this then u can substitute exp x by the above and further see if its solvable !
keep thinking :P
hint it can be done by parts
i know it can be done by parts however are you in calculus 2
well outkas do it by parts i wanna see
For convenience, 1/x is the first function.. e^x as second function.. \[\int\limits_{}^{}(\frac{1}{x}).e^x.dx = \frac{1}{x}.e^x - \int\limits_{}^{}(\frac{-1}{x^2}.e^x).dx\] \[= \frac{e^x}{x} +\int\limits_{}^{} \frac{e^x}{x^2}.dx\] But now x will move on increasing like x^3, x^4 etc etc..
yeah now you have to do integration of (e^x)/x^2 right?
yes however switch the u and dv to go back and then add them and divide by 2
Right but that will lead to x^3 then this process continues..
in otherword let dv = -1/x^2
haha outkas right :D :D
then add the integrals and divide by 2
it is a two step integral
@Omniscience poor you :P
outkas which calculus you doing
I'm don with calculus, I'm in Differential Equations right now
im good at D.E's :D
however i don't remember hardly anything from calc 3.. and i wouldn't remember the series i it weren't for review
calc 3 sucks!
indeed it does
D.E are real good, in which uni you are and which year?
nothing like drawing 3d objects on a 2d piece of paper
i'm a at a community and it's around my second/3rd year
i just cleared the first year only :/
now i will go to second
good try mukushla :)
tnx
there was a little mistake in my answer \[\int\limits \frac{e^x}{x}dx=\int\limits \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^n}{n!}dx=\sum_{n=0}^{\infty} \int\limits \frac{x^{n-1}}{n!} dx=\ln x+\sum_{n=1}^{\infty} \frac{x^{n}}{n.n!}\]

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