Having trouble with a fairly simple integration

1. Feb 21, 2012

Pandabasher

1. The problem statement, all variables and given/known data
I'm in the middle of solving a D.E using integration factors, and I've ended up with an integral on the R.H.S which is as follows: $$\displaystyle \int \frac {e^ \frac {-1}{x}}{x^4} dx$$

I've tried to use integration by parts to do it, but I keep having to use it again and again. I've checked that my integrating factor is correct, and it is. I'm just having a hard time doing it. Any help is greatly appreciated. Thanks, Matt.

2. Feb 21, 2012

Staff: Mentor

Try splitting it this way:
u = x^(-2), dv = x^(-2)e^(-1/x)dx

I haven't worked it through, so can't guarantee this will work, but this is what I would try.

3. Feb 21, 2012

Pandabasher

Okay, thanks. I'll give it a go.

4. Feb 21, 2012

dextercioby

If you let 1/x = t (new variable), what do you get ?

5. Feb 21, 2012

Dick

I would start by substituting u=(-1/x). The integration by parts should be straightforward. You only need to do it twice. Can you show where you are having problems?

6. Feb 21, 2012

Pandabasher

Using $$\displaystyle u= \frac {-1}{x}$$ I get $$\displaystyle \int e^{u} u^{2} du$$

Then I used integration by parts to get that equal to $$\displaystyle u^{2}e^{u} - \int 2u e^{u} du$$

Then using integration by parts again I get that equal to $$\displaystyle 2u e^{u} - \int 2e^{u}$$ which is just $$\displaystyle 2u e^{u} - 2e^{u}$$

Then subbing back in and taking out a factor of $$\displaystyle e^u$$ I get the answer to the first integral as $$\displaystyle e^{u}(u^{2}-2u+2)$$

And subbing u back in I get $$\displaystyle e^{ \frac {-1}{x}}( { \frac {1}{x^{2}}} + { \frac {2}{x}} + 2)$$

Does that look right?

7. Feb 21, 2012

Staff: Mentor

You can check yourself - differentiate what you ended with and you should get the integrand of your integral.

8. Feb 21, 2012

Pandabasher

Yeah I checked it with wolfram, its correct. Thanks for the tip, I'm not that good with integrals so I probably wouldn't have spotted that.

9. Mar 4, 2012

dextercioby

Late I am, but for the record, don't leave out the integration constant, because it shows what you're looking for: the 'family' of antiderivatives.