# Integral Trouble

1. Aug 27, 2005

### laker88116

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

So far I know that I need to use integration by parts, I let $u= ln(2x+1)$ and so $du= \frac {dx}{2x+1}$. Also, I said $dv= dx$ and $v=x$.

So then plugging this into the equation for integration I get:

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

Then I determine that I need to do integration by parts again on the latter half of the function. So, for $\int \frac {2x}{2x+1}dx$, I let $u= 2x$ and so $du= xdx$. Also, I said $dv= \frac {dx}{2x+1}$ and $v= \frac {ln(2x+1)}{2}$.

So then plugging this into the equation for integration I get:

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

Now, I have like terms so I say that:

$$\int ln(2x+1)dx = xln(2x+1) - [xln(2x+1) - \int ln(2x+1)dx].$$

I am not sure where I made an error here. Any help is appreciated.

2. Aug 27, 2005

### Galileo

Oi, so complicated >_<

You don't have to integrate by parts again. You can split the fraction like:

$$\frac{2x}{2x+1}=1-\frac{1}{2x+1}$$

But even simpler is first solving:

$$\int \ln x dx$$
then the integral should be a piece of cake. The straight way to Rome is not always the shortest, nor the easiest to follow.

3. Aug 27, 2005

### iNCREDiBLE

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

4. Aug 27, 2005

### laker88116

Wow, how did I miss that, thanks so much.

5. Aug 27, 2005

### Hurkyl

Staff Emeritus
Anyways, as to the actual method you implemented, your second integration was simply undoing the first integration by parts you tried. That's why you got a useless result at the end.

6. Aug 27, 2005

### Hurkyl

Staff Emeritus
It's not productive to do the problem for the person asking for help. (especially when you do it wrong!) Fortunately, the poster had already figured it out from the hints!