# Math Help - [SOLVED] Integration and natural logs

1. ## [SOLVED] Integration and natural logs

I have run into a brick wall that is driving me insane (see following picture: )

It seems simple, yet I'm not getting the answer in the book, but rather I'm looping through the Integration By Parts method over and over again.

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

I set:

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

and get:

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

and from there I integrate by partial fractions:

$2\int{x\frac{1}{2x+1}}$

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

then I get:

$2[\frac{x^2}{2(2x+1)}-\int{\frac{x^2}{(2x+1)^2}}]$

I might be wrong, but it seems like an endless loop to me

Could anyone help me out?

2. Hello,

If you integrate by parts once, then integrate by parts with the reverse substitution, it is sure you won't get to the answer

After the first integration by parts, you have :

$\int \frac{2x}{2x+1} ~dx$

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

Can you do it ?

3. Use the sub: $u = 2x + 1 \ \Rightarrow \ du = 2 dx \iff \frac{du}{2} = dx$

So: $\int \ln (2x + 1) \ dx \ = \ \frac{1}{2} \int \ln u \ du$

which is a standard one.

4. $\int{\frac{2x}{2x+1}}$
Why use integration by parts for this one? Simply add and subtract one to get:
$\int (1 - \frac{1}{2x+1})$

It should be simple enough now.

5. Originally Posted by o_O
which is a standard one.
Standard one that you get with an integration by parts. It may not sound as standard as you know it

6. Originally Posted by Moo
Hello,

If you integrate by parts once, then integrate by parts with the reverse substitution, it is sure you won't get to the answer

After the first integration by parts, you have :

$\int \frac{2x}{2x+1} ~dx$

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

Can you do it ?
Hmm, where did you get the $-1$ from on the first part?

7. Originally Posted by Coco87
Hmm, where did you get the $-1$ from on the first part?
1-1=0

I made "appear" the denominator in order to simplify. It is a similar process to a polynomial division.

8. Originally Posted by Moo
1-1=0

I made "appear" the denominator in order to simplify. It is a similar process to a polynomial division.
Ah, thanks! , I didn't know that is allowed.

I'm still having an issue though

$(x-\frac{1}{2})\ln{(2x+1)}-x+C$

However, the answer in the back of the book is:

$\frac{1}{2}(2x+1)\ln{(2x+1)}-x+C$

I'm not sure where I messed up

9. Originally Posted by Coco87
Ah, thanks! , I didn't know that is allowed.

I'm still having an issue though

$(x-\frac{1}{2})\ln{(2x+1)}-x+C$

However, the answer in the back of the book is:

$\frac{1}{2}(2x+1)\ln{(2x+1)}-x+C$

I'm not sure where I messed up

$\frac 12 (2x-1)=\frac 12 (2x)-\frac 12=\dots ??$

and by the way, it must be the absolute value of 2x+1 in the logarithm, since it is not defined over non-positive numbers.

10. When you did integration by parts you subtract the integral ( $-\int vdu$. You broke up that integral into two parts. Did you distribute the negative sign to both?

After that change of sign, they just brought out a constant to make it look neater.

11. I see how they connect now, I'm still unsure where they got the 2x from to multiply by

12. $(x+\frac{1}{2}) = \frac{1}{2}(2x+1)$

...

13. Ah, now I got ya! I was looking at it the wrong way I guess (long day).

Thanks for helping me out guys!