Solving Integral Trouble: \int ln(2x+1)dx

[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.

 

[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.

 

[iNCREDiBLE]

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

 

[laker88116]

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

 

[Hurkyl]

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.

 

[Hurkyl]

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!