• Support PF! Buy your school textbooks, materials and every day products Here!

Using Integration by parts

  • #1
120
0

Homework Statement



∫[itex]\frac{1}{x^{2}*ln(x)}[/itex]

Homework Equations



∫udv = uv-∫vdu

u=ln(x)
du = [itex]\frac{1}{x}[/itex]dx
dv = [itex]x^{2}[/itex]dx
v = [itex]\frac{x^{3}}{3}[/itex]

The Attempt at a Solution



Using the above formula I got [itex]\frac{x^{3}}{3}[/itex]*ln(x) - [itex]\frac{x^{3}}{9}[/itex] + C

Am I doing this correctly or do I have to input u= [itex]\frac{1}{ln(x)}[/itex] and dv = [itex]\frac{1}{x^{2}}[/itex]
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,302
998

Homework Statement



∫[itex]\frac{1}{x^{2}*ln(x)}[/itex]

Homework Equations



∫udv = uv-∫vdu

u=ln(x)
du = [itex]\frac{1}{x}[/itex]dx
dv = [itex]x^{2}[/itex]dx
v = [itex]\frac{x^{3}}{3}[/itex]

The Attempt at a Solution



Using the above formula I got [itex]\frac{x^{3}}{3}[/itex]*ln(x) - [itex]\frac{x^{3}}{9}[/itex] + C

Am I doing this correctly or do I have to input u= [itex]\frac{1}{ln(x)}[/itex] and dv = [itex]\frac{1}{x^{2}}[/itex]
First of all, you should include a dx with your integral, [itex]\displaystyle \int \frac{dx}{x^{2}\ln(x)}\ .[/itex]

Your dv is wrong. It should be [itex]\displaystyle dv=\frac{1}{x^2}\,dx\ .[/itex]
.
 
  • #3
631
0
First of all, you should include a dx with your integral, [itex]\displaystyle \int \frac{dx}{x^{2}\ln(x)}\ .[/itex]

Your dv is wrong. It should be [itex]\displaystyle dv=\frac{1}{x^2}\,dx\ .[/itex]
.
Follow this with a substitution: y = f(x). Think what f(x) should be so that your equations take a simpler form.
 
  • #4
120
0
First of all, you should include a dx with your integral, [itex]\displaystyle \int \frac{dx}{x^{2}\ln(x)}\ .[/itex]

Your dv is wrong. It should be [itex]\displaystyle dv=\frac{1}{x^2}\,dx\ .[/itex]
.
I don't understand, why is [itex]\displaystyle dv=\frac{1}{x^2}\,dx\ .[/itex] and u = ln(x) and not u = [itex]\frac{1}{ln(x)}[/itex]

Follow this with a substitution: y = f(x). Think what f(x) should be so that your equations take a simpler form.
I'm sorry, I don't understand what this means.
 
  • #5
30
1
I don't understand, why is [itex]\displaystyle dv=\frac{1}{x^2}\,dx\ .[/itex] and u = ln(x) and not u = [itex]\frac{1}{ln(x)}[/itex]



I'm sorry, I don't understand what this means.
[itex]u =\frac{1}{ln(x)}[/itex] and [itex]dv=\frac{1}{x^2}\,dx\ [/itex] are the correct forms. The reason for this is that integration by parts is ∫udv = uv-∫vdu. Thus, when you multiply u and dv together it has to come out to the original integrand.
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,302
998
I don't understand, why is [itex]\displaystyle dv=\frac{1}{x^2}\,dx\ .[/itex] and u = ln(x) and not u = [itex]\frac{1}{ln(x)}[/itex]
Well, you're right that you would also need [itex]\displaystyle u=\frac{1}{\ln(x)} .[/itex]

But that scheme for integration by parts was yours.

Simple substitution seems to do better, but this integral cannot expressed in terms of elementary functions.
 
  • #7
120
0
Well, you're right that you would also need [itex]\displaystyle u=\frac{1}{\ln(x)} .[/itex]

But that scheme for integration by parts was yours.

Simple substitution seems to do better, but this integral cannot expressed in terms of elementary functions.
Are you saying that there is an easier way to do this?
 
  • #8
30
1
Are you saying that there is an easier way to do this?
I think he's saying that the solution is messy because you can't write the solution in terms of elementary functions (i.e. finite number of terms using exponents, trig functions, constants, etc).
 
  • #9
120
0
I'm not sure this can be solved.

with:

u = [itex]\frac{1}{ln(x)}[/itex]
du = -[itex]\frac{1}{xln^{2}(x)}[/itex] dx
dv = [itex]\frac{1}{x^{2}}[/itex]dx
v = -[itex]\frac{1}{x}[/itex]

i get: -[itex]\frac{1}{xln(x)}[/itex] - ∫-[itex]\frac{1}{x}[/itex] * [itex]\frac{1}{xln^{2}(x)}[/itex] dx

so then I would have to do integration by parts again because I have x * the natural log

this seems like it would go on for a while.

edit: Am I doing this correctly?
 
Last edited:
  • #10
631
0
Hint: In the original integral, substitute y = 1/x.

You will then be able to reduce this to a standard integral. By standard integral, I mean an integral which cannot be reduced any further.
 
  • #11
120
0
∫[itex]\frac{dx}{x^{2}*ln(x)}[/itex]

I'm afraid I don't understand what you mean by that.

I dont' see a y in this equation.

Do you mean make [itex]\frac{1}{x}[/itex] = y ?
 
  • #12
631
0
  • #13
120
0
∫[itex]\frac{y^{2}dx}{ln(x)}[/itex]

Would i use integration by parts now?
 
  • #14
631
0
∫[itex]\frac{y^{2}dx}{ln(x)}[/itex]

Would i use integration by parts now?
You need to convert all x's to y's. That means:
1. ln(x) becomes some function of y (after replacing 1/x by y),
2. dx becomes some function*dy (after replacing 1/x by y).

Please look at the link I sent in previous reply to see an example of how this is done. I'll repeat it here: http://en.wikipedia.org/wiki/Integration_by_substitution#Examples
 
  • #15
30
1
So you have y=1/x. When you take the derivative, you get dy/dx=-1/x^2, which can be written as dy=-dx/x^2. Then substitute for all x's.
 
  • #16
120
0
y = 1/x

dy = -[itex]\frac{1}{x^{2}}[/itex]dx

dx = [itex]\frac{-1}{y^{2}}[/itex]dy


∫[itex]\frac{y^{2}\frac{-1}{y^{2}}dy}{ln(\frac{1}{y})}[/itex]


∫[itex]\frac{dy}{ln\frac{1}{y}}[/itex]

well i replaced everything and now if I do integration by parts It won't end well because I can't integrate 1/ln, and if I integrate dy i'll end up with y in the integral again.
 
  • #17
631
0
dx = [itex]\frac{-1}{y^{2}}[/itex]dy
∫[itex]\frac{y^{2}\frac{-1}{y^{2}}dy}{ln(\frac{1}{y})}[/itex]


∫[itex]\frac{dy}{ln\frac{1}{y}}[/itex]
You forgot a minus sign in the second expression there. Also, do you think you can relate ln(1/y) and ln(y)? This will reduce your expression to a 'standard integral'.

well i replaced everything and now if I do integration by parts It won't end well because I can't integrate 1/ln, and if I integrate dy i'll end up with y in the integral again.
As I said before, you will end up with a 'standard integral', an integral which *cannot* be reduced further. It is known as the 'Logarithmic Integral'.
 
  • #18
120
0
I haven't learned about logarithmic integrals. Do you think I should be using a technique my teacher hasn't taught?

This whole problem started out as a separable differential equation. My teacher did the left hand side and asked us to do the right hand side which was the integral I posted.
 
  • #19
631
0
I haven't learned about logarithmic integrals. Do you think I should be using a technique my teacher hasn't taught?
Here you go then: http://mathworld.wolfram.com/LogarithmicIntegral.html

This whole problem started out as a separable differential equation. My teacher did the left hand side and asked us to do the right hand side which was the integral I posted.
It might be possible that you did a mistake before arriving at the integral (so the integrand is wrong). Otherwise, maybe this is the solution.
 
  • #20
120
0
He wrote the integral for us to solve.

Thanks for the link, I don't think I'll have time to read it now because my class starts in 6 hours and I think I should sleep before then. Should I just turn in what i've done so far even though it's wrong?
 
  • #21
631
0
Turn in the final expression you got, following the suggestions here. It's not wrong.

Maybe he didn't write it down correctly, or maybe he wanted you to look up Logarithmic Integral by yourself.
 
  • #22
120
0
Alright, thanks for the help!
 
  • #23
631
0
You're welcome!
 

Related Threads on Using Integration by parts

  • Last Post
Replies
5
Views
3K
Replies
14
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
6
Views
22K
Replies
2
Views
1K
Replies
8
Views
1K
Replies
2
Views
1K
Top