Using Integration by parts

1. Sep 6, 2012

november1992

1. The problem statement, all variables and given/known data

∫$\frac{1}{x^{2}*ln(x)}$

2. Relevant equations

∫udv = uv-∫vdu

u=ln(x)
du = $\frac{1}{x}$dx
dv = $x^{2}$dx
v = $\frac{x^{3}}{3}$

3. The attempt at a solution

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

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

2. Sep 6, 2012

SammyS

Staff Emeritus
First of all, you should include a dx with your integral, $\displaystyle \int \frac{dx}{x^{2}\ln(x)}\ .$

Your dv is wrong. It should be $\displaystyle dv=\frac{1}{x^2}\,dx\ .$
.

3. Sep 6, 2012

Sourabh N

Follow this with a substitution: y = f(x). Think what f(x) should be so that your equations take a simpler form.

4. Sep 6, 2012

november1992

I don't understand, why is $\displaystyle dv=\frac{1}{x^2}\,dx\ .$ and u = ln(x) and not u = $\frac{1}{ln(x)}$

I'm sorry, I don't understand what this means.

5. Sep 6, 2012

$u =\frac{1}{ln(x)}$ and $dv=\frac{1}{x^2}\,dx\$ 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. Sep 7, 2012

SammyS

Staff Emeritus
Well, you're right that you would also need $\displaystyle u=\frac{1}{\ln(x)} .$

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. Sep 7, 2012

november1992

Are you saying that there is an easier way to do this?

8. Sep 7, 2012

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. Sep 7, 2012

november1992

I'm not sure this can be solved.

with:

u = $\frac{1}{ln(x)}$
du = -$\frac{1}{xln^{2}(x)}$ dx
dv = $\frac{1}{x^{2}}$dx
v = -$\frac{1}{x}$

i get: -$\frac{1}{xln(x)}$ - ∫-$\frac{1}{x}$ * $\frac{1}{xln^{2}(x)}$ 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: Sep 7, 2012
10. Sep 7, 2012

Sourabh N

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. Sep 7, 2012

november1992

∫$\frac{dx}{x^{2}*ln(x)}$

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

I dont' see a y in this equation.

Do you mean make $\frac{1}{x}$ = y ?

12. Sep 7, 2012

Sourabh N

Yes! I don't know if you are familiar with the method of substitution, so have a look. It's a very useful technique

13. Sep 7, 2012

november1992

∫$\frac{y^{2}dx}{ln(x)}$

Would i use integration by parts now?

14. Sep 7, 2012

Sourabh N

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. Sep 7, 2012

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. Sep 7, 2012

november1992

y = 1/x

dy = -$\frac{1}{x^{2}}$dx

dx = $\frac{-1}{y^{2}}$dy

∫$\frac{y^{2}\frac{-1}{y^{2}}dy}{ln(\frac{1}{y})}$

∫$\frac{dy}{ln\frac{1}{y}}$

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. Sep 7, 2012

Sourabh N

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'.

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. Sep 7, 2012

november1992

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. Sep 7, 2012

Sourabh N

Here you go then: http://mathworld.wolfram.com/LogarithmicIntegral.html

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. Sep 7, 2012

november1992

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?