# Integrate by parts

1. Mar 20, 2006

### goel_heena

Hi

Can anyone pls suggest the trick to do integration by parts such as:

Intergration {(1/x) (e^-cx) }dx.

Which function normally we should take as first and second function.
Is there a rule to decide on it.

thanks

2. Mar 20, 2006

### d_leet

It takes a bit of practice to start noticing which function you should chose to be the one you differentiate or integrate, for the one you have I think it might be better to integrate 1/x and differentiate e, but I'm not completely sure how well that will work.

3. Mar 20, 2006

### HallsofIvy

Here's a good rule: Try one and if it doesn't work, try the other! Surely that would have taken less time than waiting for a reply here.

4. Mar 20, 2006

### VietDao29

Sorry to interrupt you guys, but can this expression be done by Integration by Parts?
It seems that the answer does contains the Exponential Integral. Am I missing something?
$$\int x e ^ x dx$$ or something along those lines, I don't think Integration by Parts will work...
Am I missing something?

5. Mar 20, 2006

### Hootenanny

Staff Emeritus
You can integrate an exponent;

$$\int e^{kx} = \frac{1}{k}e^{kx} + C$$

That's if I remember correctly...

6. Mar 20, 2006

### VietDao29

Yes, of course, you can integrate exponents (and you seem to remember it correctly , just one note, you forgot the variable of integration ). However, just look at the first post again, his integral is:
$$\int \frac{e ^ {-cx}}{x} dx$$, and not
$$\int e^{kx} dx$$ :)
The Exponential Integral is a function which is like Sine Integral or Cosine Integral.
You may want to take a brief look at this page. He may also integrate that expression using series, but it's not going to give a nice result...
And I just wonder if it can be done with Integration by Parts.

Last edited: Mar 20, 2006
7. Mar 20, 2006

### goel_heena

Yes its read as $$\int x e ^ x dx$$ ...

8. Mar 20, 2006

### Hootenanny

Staff Emeritus
Something like this perhaps?

$$\int \; u\frac{dv}{dx} = uv - \int \; v\frac{du}{dx}$$

So if we let $u = \frac{1}{x}$ and $\frac{dv}{dx} = e^{-cx}$ gives;

$$= \frac{1}{x}\cdot\frac{-1}{c}e^{-cx} - \int\frac{-1}{c}e^{-cx}\cdot\frac{-1}{x^2}\;\;dx$$

I won't carry on because we're not suppose to show full solutions, but by the looks of it, it can be done...

9. Mar 20, 2006

### VietDao29

Err???
There's some huge differences from your first post. In your first post, you said:
and now you say it's Int xexdx... As far as I can tell, there's absolutely no similarity between the two...
goel_heena, it's important to post the problem exactly.
------------------
by the way, have you tried anything?
Now if you choose u = ex, dv = xdx, then you'll have:
$$\int x e ^ x dx = \frac{1}{2} x ^ 2 e ^ x - \frac{1}{2} \int x ^ 2 e ^ x dx$$. Arrghh, it looks even more complicated then your original integral, so let's try another way.
This time, what should be u, and what should be dv, you think?
Can you go from here? :)

10. Mar 20, 2006

### VietDao29

No offense, but are you sure that's the right formula for Integration by Parts? Where is your variable of integraion??? Please, make sure that you are sure about something before helping someone else, or you may just make them more confused. :grumpy:
The correct version:
$$\int u dv = uv - \int vdu$$.
IMO, that looks way harder than the original integral.

11. Mar 20, 2006

### Hootenanny

Staff Emeritus
Wait a minute! What is your question

It is (1)
$$\int \; xe^{-cx} \;\; dx$$

or it it (2)
$$\int \; \frac{e^{-cx}}{x} \;\; dx$$

???

As VietDao said, it is very important to accurately present your question. Either equation is solvable using the integration by parts.

12. Mar 20, 2006

### Hootenanny

Staff Emeritus
Yes I do know about it. It is the correct formula, it is equivalent to your formula. My variable of integration is dx.

13. Mar 20, 2006

### Hootenanny

Staff Emeritus
Looking at it now the better subsitution could be

$$u = e^{-cx}$$

$$\frac{dv}{dx} = \frac{1}{x}$$

I don't know, I haven't worked through the problem, but this one looks like the most promising

14. Mar 20, 2006

### VietDao29

Are you sure the second one is solvable?
Hmm, one last advice, please work it out first, before making such assumption.
And as I pointed out in my previous post, I don't see it very promising. Don't be surprised if you keep getting monstrous expressions by using Integration by Parts. :)

Last edited: Mar 20, 2006
15. Mar 20, 2006

### Hootenanny

Staff Emeritus
Point taken...