Calculating Integral of e^x/x - Find My Mistake

[3.1415926535]

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 definitely wrong Anyway, have a look and tell me where is my mistake
http://docs.google.com/View?id=dd4jpgg_1g8ztn3g5

 

[HallsofIvy]

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.

 

[3.1415926535]

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]

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.

 

[The Chaz]

How about googledocs?

 

[n.karthick]

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!

 

[3.1415926535]

Here is the google docs equation
http://docs.google.com/View?id=dd4jpgg_1g8ztn3g5

 

[3.1415926535]

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

 

[n.karthick]

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.

 

[losiu99]

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

 

[n.karthick]

Whatever way I try it is bouncing back to beginning.
Finally wikipedia page gave me the answer.
http://en.wikipedia.org/wiki/Exponential_integral

 

[3.1415926535]

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|)

 

[3.1415926535]

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

 

[The Chaz]

Whatever way I try it is bouncing back to beginning.
Finally wikipedia page gave me the answer.
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)

 

[tony.c.tan]

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

 

[Anonymous217]

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