Using Integration by parts

In summary: 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.
  • #1
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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]
 
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  • #2
november1992 said:

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
SammyS said:
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
SammyS said:
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]

Sourabh N said:
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
november1992 said:
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
november1992 said:
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
SammyS said:
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
november1992 said:
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
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
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
∫[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
november1992 said:
Do you mean make [itex]\frac{1}{x}[/itex] = y ?

Yes! I don't know if you are familiar with the method of substitution, so have a look. It's a very useful technique :smile:
 
  • #13
∫[itex]\frac{y^{2}dx}{ln(x)}[/itex]

Would i use integration by parts now?
 
  • #14
november1992 said:
∫[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
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
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
november1992 said:
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
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
november1992 said:
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
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
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
Alright, thanks for the help!
 
  • #23
You're welcome!
 

1. How do you know when to use integration by parts?

Integration by parts is typically used when you have a product of functions in the integrand, such as a polynomial times a trigonometric function or an exponential function. It is also useful when you have an integral that is difficult to evaluate using other methods, such as substitution or partial fractions.

2. What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are functions of x and dv and du are their respective derivatives with respect to x.

3. What is the process for using integration by parts?

The process for using integration by parts is as follows:1. Identify u and dv in the integrand.2. Differentiate u to get du and integrate dv to get v.3. Plug u, du, v, and dv into the integration by parts formula to obtain the new integral.4. Solve for the new integral, which should be simpler than the original one.5. If necessary, use integration by parts again on the new integral to simplify it further.

4. What is the purpose of using integration by parts?

The purpose of using integration by parts is to simplify integrals that are difficult to evaluate using other methods. It can also help to break down complicated integrals into smaller, more manageable parts.

5. Are there any special cases where integration by parts is particularly useful?

Yes, there are a few special cases where integration by parts is particularly useful. One example is when the integrand contains logarithmic functions. Another is when the integrand contains inverse trigonometric functions. In these cases, choosing u and dv strategically can simplify the integral significantly.

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