Integration by part question

1. Sep 26, 2011

QuarkCharmer

1. The problem statement, all variables and given/known data
$$\int ln(2x+1)dx$$

2. Relevant equations

3. The attempt at a solution

$$\int ln(2x+1)dx$$

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

$$\int ln(2x+1)dx = xln(2x+1)-2\int \frac{x}{2x+1}dx$$

And there is where I am stuck. I can't use trig substitution either. I don't know how to integrate x/(2x+1). I tried successive integrations by part, they didn't work. I tried u-substitution, no luck. Wolfram alpha wasn't any help in solving this either. I know there must be a way for my to solve it using only substitution and integration by part. Can someone just point me in the right direction please?

2. Sep 26, 2011

flyingpig

I recommend you do

u = 2x + 1
du = 2dx

New integral becomes

$$\frac{1}{2}\int \ln(u) du$$

Now do integration by parts

3. Sep 26, 2011

QuarkCharmer

But there is an x on top of that integral?

$$\int \frac{x}{2x+1} dx$$
u = 2x+1
$$\int \frac{x}{2u}du$$

Then I can't really do anything with that x? Or you mean integrate by parts now?

4. Sep 26, 2011

flyingpig

Let's try to use a different symbol...

$$\alpha = 2x + 1$$
$$d\alpha = 2dx$$

New integral becomes

$$\frac{1}{2}\int \ln(\alpha) d\alpha$$

5. Sep 26, 2011

Staff: Mentor

Which is essentially the integral QuarkCharmer started with.

A simpler way to go is to do polynomial long division in x/(2x + 1) to get 1/2 - (1/2)/(2x + 1). This results in a much simpler integral.

6. Sep 26, 2011

flyingpig

Yeah, but you don't have to deal with the fractions

7. Sep 26, 2011

QuarkCharmer

I started with flyingpigs idea..

$$Y = \int ln(2x+1) dx$$

$$\int ln(2x+1) dx = xln(2x+1) - 2 \int \frac{x}{2x+1}$$

Let Z = 2x+1
So that dZ/dx = 2

$$\int ln(2x+1) dx = xln(2x+1) - 2 (\int \frac{x}{Z})$$

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

So now I have:
$$\int ln(2x+1) dx = xln(2x+1) - 2(\frac{1}{2} xln(Z)- \frac{1}{2} \int ln(Z)dx )$$
Now I stick back in for Z:
$$\int ln(2x+1) dx = xln(2x+1) - 2(\frac{1}{2} xln(2x+1)- \frac{1}{2} \int ln(2x+1)dx )$$

Now, I think that since the initial problem was the integral of ln(2x+1), and in the beginning I let Y = the initial problem, I can say that:

$$Y = \int ln(2x+1) dx = xln(2x+1) - 2(\frac{1}{2} xln(2x+1)- \frac{1}{2} Y)$$

$$Y = \int ln(2x+1) dx = xln(2x+1) - xln(2x+1) + Y)$$ /distributing the 2 through

Awe wait, that's going to make Y = Y....

Can you please explain both of your methods? I don't see how polynomial long division will work here, nor can I see the substitution flyingpig is talking about. Forgive my ignorance.

8. Sep 26, 2011

Staff: Mentor

To finish this off, use polynomial long division to divide x by 2x + 1. When I do this, I get 1/2 - (1/2)/(2x + 1).

This means that the last integral above is
$$-2\int \frac{x}{2x+1}dx = \int \left(-1 + \frac{1}{2x + 1}\right)dx$$

Split the integral on the right into two. One of them is really easy, and the other can be done using an ordinary substitution.

If you don't know how to do polynomial long division, I believe there's a page on Wikipedia that shows how and gives some examples.

9. Sep 26, 2011

Staff: Mentor

I would much rather deal with fractions than with a ponderous integration by parts that doesn't seem to go anywhere.

10. Sep 26, 2011

flyingpig

But it's just ln(x) as the integrand.

@QuickCharmer, my integral involving alpha is a "new" integral

11. Sep 26, 2011

Staff: Mentor

Actually, the integrand is ln(2x + 1), from what you have in post #4, and this is the same as what you started with.

12. Sep 26, 2011

flyingpig

Yeah but the 2x + 1 thing and what comes after is bothering Quark. He looks proficient with by parts.

I've got a good feeling on this one

13. Sep 26, 2011

romsofia

Just start by integrating ln(x), use integration by parts on that. Once you do that, you'll see this problem becomes much easier, and you'll be able to use the substitution u=2x+1.

14. Sep 27, 2011

Jxs63J

Others above may have said this, but let u = 2x + 1. Then x = (u - 1) / 2. You then get 2 easy integrals.

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