Question

Integration by parts involving Inx:

http://i1220.photobucket.com/albums/dd441/smileesue/math-1.jpg

How would you break this integral into two pieces? would it be where the subtraction is? I just don't how to partially integrate it since I don't know which values to assign as u and dv...

Answer 1

separate in to two partial fractions.

1/lnx^2 + 1/lnx

Answer 2

\[\int\limits_{e}^{e ^{2}}(1-Inx)/(Inx)^2 dx = \int\limits_{e}^{e ^{2}}1/(Inx)^2dx - \int\limits_{e}^{e ^{2}}(1/Inx) dx\]
That's all I got to.

Answer 3

I tried u = 1/Inx but I can't go on further..

Answer 4

i guess i don't know too ... seems like worth trying.

Answer 5

ugly
http://upload.wikimedia.org/wikipedia/en/math/a/3/9/a39117dd6038c21b11c90a777c1f7699.png

Answer 6

Have you tried u = lnx?
Then, x = e^u and du = (1/x)dx.
The integral becomes:
\[\int\limits_{a}^{b} (e^u/u^2 - e^u/u)du\]
If I didn't make any careless mistake

Answer 7

And that should be an indefinite integral, a and b are arbitrary limits.

Answer 8

there might be some other elegant method ... changing limits.

Answer 9

http://www.wolframalpha.com/input/?i=Integral+%28%28e%5Ex%29%2F%28x%5E2%29+-+%28e%5Ex%29%2Fx%29
It works :-) haha

Answer 10

In this case, x = lnx, so it becomes -x/lnx

Answer 11

So would you have to partially integrate \[\int\limits_{}^{}(e ^{u}/u ^{2})du - \int\limits_{}^{}(e ^{u}/u)du\] for the respective terms? So for the first term, u=e^u and dv=(1/u^2)du? I am slightly confused

Answer 12

Indeed. But I am unsure that this is the right way. The integrals don't look simplified. I was just amazed that what I did work, haha.

Answer 13

this is getting crazy ... looks something like gamma function

Answer 14

http://en.wikipedia.org/wiki/Gamma_function
if the limits were 0 to inf then we'll be able to get the values easily.

Answer 15

Gaah, was typing the reply and the site crashed. Anyway, I think that both integrals are easy. You can do it by definition, i.e., integral of (e^u)/u = exponential integration of u + a constant.

Answer 16

Or maybe not. Yeah, you would have to use the definition if you went through my path. e^u/u is not an elementary function... There has to be another way.

Answer 17

This was what was written below the question. I'm not sure if that'd be a big help though..

Answer 18

What bothers me is that with that substitution (u = 1/lnx), it seems that there is a du factor missing. I am committing a silly mistake somewhere, but my integral keeps getting reduced to u + du times some factor.

Answer 19

Oh I see. I'll put this problem aside for now and try again. Thanks bmp and experimentX for your help though :)

Answer 20

Got it!

Answer 21

Damn, lost my answer.

Answer 22

Anyway, simplify the expression by dividing it by (lnx)^2. Multiply it out by (x/x), then apply the substitution hinted above. We get two integrals, one of them:
integral of udx and the other integral of xdu. Then, do the xdu integral by parts, we will have:
ux - integral udx, that will cancel out the other integral.
So we are left with u*x, that is, x/lnx, QED

Answer 23

-x/lnx*

Answer 24

by dividing it by (lnx)^2, do you mean the original expression? Or do you mean multiplying? Because then wouldn't the denominator become (Inx)^4?

Answer 25

No, I mean just simplifying it to obtain:
\[\int\limits_{e}^{e^2} (1/lnx^2 + 1/lnx)dx\]

Answer 26

Hope the font is not too small

Answer 27

Oh I see. Thank you! :)