anonymous
  • anonymous
Find the indefinite integral of: [e^x]/[x] dx I tried using parts, but it seemed to only get worse.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1333584024637:dw|
JamesJ
  • JamesJ
This integral doesn't have a closed form in terms of elementary functions. The easiest way to deal with this if you want to get at least something is to write down and integrate the power series.
anonymous
  • anonymous
So you are saying that it is not solvable via integration by parts, substitution, partial fraction decomposition?

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JamesJ
  • JamesJ
Nope, it's not.
anonymous
  • anonymous
Ok wait Let me show you the original integral. I actually split this up:
anonymous
  • anonymous
|dw:1333584297934:dw|
anonymous
  • anonymous
The textbook says: The following exercises are all elementary functions. So I can guarantee you that this one is, and it falls for this criteria, according to the text.
JamesJ
  • JamesJ
Notice this is \[ \int \left( \frac{1}{x} - \frac{1}{x^2} \right) e^x \ dx \] Now ... integrate that second term by parts \( -e^x/x^2 \) and something magical will happen.
anonymous
  • anonymous
How about: Let e^x = |dw:1333584476899:dw|
anonymous
  • anonymous
And then I simplify?
JamesJ
  • JamesJ
No... \[ \int \frac{-1}{x^2} e^x \ dx = \frac{1}{x} e^x - \int \frac{1}{x} e^x \ dx \]
anonymous
  • anonymous
Wow!
anonymous
  • anonymous
Ok I will follow your advice. I will let you know. You can go for now if you would like,
anonymous
  • anonymous
I will call you by "Good Answer"
JamesJ
  • JamesJ
Well, do you see what do now?
JamesJ
  • JamesJ
Your integral is \[ \int \left(\frac{1}{x} - \frac{1}{x^2} \right) e^x \ dx = \int \frac{1}{x} e^x \ dx + \int \frac{-1}{x^2} e^x \ dx \] We just evaluated that second integral as above. Hence that entire expression is just equal to \[ \int \frac{1}{x} e^x \ dx + \frac{1}{x} e^x - \int \frac{1}{x} e^x \ dx = \frac{1}{x} e^x \]
anonymous
  • anonymous
I see. Ok next time I will experiment more with the integrand. Thank you.
JamesJ
  • JamesJ
This is an example of the following general principle. If \[ g(x) = f(x)e^x \] then \[ g' = (f + f')e^x \] In your example, f(x) = 1/x and therefore f'(x) = -1/x^2 Now you've seen this pattern you should recognize it immediately in the future. Here's another example ...
JamesJ
  • JamesJ
\[ \int (\cos x - \sin x)e^x \ dx \]
JamesJ
  • JamesJ
You could torture yourself using integration by parts to solve it. But you instead, you should now be able to see by inspection the integrand is just the derivative of \[ \cos x . e^x \]
JamesJ
  • JamesJ
This generalizes further to functions of the sort \[ g(x) = f(x)e^{ax} \] Hence the integral \[ \int \frac{x-1}{x^3} e^{2x} \ dx \] should now also be accessible to you. This one is a little more subtle. See if you can solve it.
anonymous
  • anonymous
Thank you James. I really appreciate you disclosing this trend.

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