# Integral of e^x/x

First of all, I want to clarify that i know the answer is Ei(x). I have found a way to calculate this integral but the result is definetely wrong Anyway, have a look and tell me where is my mistake

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HallsofIvy
Homework Helper
Many people (and I am one when I am not using my home computer where I know how strong my virus- protection is) will not open "word" files. They are notorious for harboring viruses.

Many people (and I am one when I am not using my home computer where I know how strong my virus- protection is) will not open "word" files. They are notorious for harboring viruses.

Really ? Didn't know that... Well, i can garentee it contains only calculus. Apart from that why why would someone upload a doc file with viruses to physics forums?? I hope i learn "latex" soon so i won't need these word documents

Mark44
Mentor
Really ? Didn't know that... Well, i can garentee it contains only calculus. Apart from that why why would someone upload a doc file with viruses to physics forums??
For at least two reasons: they knew that the file contained a virus and wanted to spread it; they didn't know the file contained a virus. For a person intent on spreading a computer virus, there is nothing sacrosanct about physicsforums.

It seems that you have missed 1/x in step 2. The final answer you got is just e^x +
c. How it can be? Can you check step no. 2?

Regarding MS word one can disable macros and defend from viruses. For me it worked!!

Here is the google docs equation

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It seems that you have missed 1/x in step 2. The final answer you got is just e^x +
c. How it can be? Can you check step no. 2?

Regarding MS word one can disable macros and defend from viruses. For me it worked!!

I know it can't be e^x+c ... I didn't miss it!! Check the google document du=1/x dx

e^lnx = x is true only if x>0 right? I tried conventional method to solve this integral and i got different answer. I have to check my answer.

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It's the 5th line. $$u$$ is independent variable here, so $$u'=1$$.

e^lnx = x is true only if x>0 right? I tried conventional method to solve this integral and i got different answer. I have to check my answer.

Of course. There can't be a ln(-|x|)

It's the 5th line. $$u$$ is independent variable here, so $$u'=1$$.

Really? I thought that since u=g(x) u'=g'(x)=lnx'=1/x

Whatever way I try it is bouncing back to beginning.
http://en.wikipedia.org/wiki/Exponential_integral
Gave you WHAT answer? That article doesn't explain where the error is...

It's the 5th line. $$u$$ is independent variable here, so $$u'=1$$.
Yes, you are having problems with the chain rule. If v = e^u, then dv/du = e^u, and dv = (e^u)du, which doesn't help solve the integral.
If you wanted to take dv/dx, then that would equal (dv/du)*(du/dx), but that's not what you're doing... (and won't help any)

Try this one:
$$e^x\sum_{i=1}^\infty(i-1)!x^{-i}$$

$$e^{elnx} = e^x$$ ?