What is the MacLaurin series for f(x)=ln(1+x^2)?

[mateomy]

MacLaurin Series Integration...

I have to find the MacLaurin for f(x)=ln(1+x^2)

So i started off by finding the derivative of the function getting

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

My issue lies with the 2x in the numerator. I know how to bring the x into the series, but the two? Do I leave it on the outside or do I bring it in and put it to the nth power? I think my brain is fried, I seem to be getting contradictory information from various sources. So, I am bringing it here and throwing it up on the board.

Should it look like this:

$$\sum_{n=0}^{\infty} (-1)^n x^{2n+1} 2^n$$

or this:

$$2 \sum_{n=0}^{\infty} (-1)^n x^{2n+1}$$

Thanks.

 

[mateomy]

...and then I integrate it from there. I get that part, its just this one little step.

 

[HallsofIvy]

To find the MacLaurin series for
$$\frac{2x}{1+ x^2}$$
Think of it as
$$(2x)\left(\frac{1}{1-(-x^2)}$$

Now, the sum of a geometric series is given by
$$\sum ar^n= \frac{a}{1- r}$$
so think of this as a geometric series with $r= -x^2$ and a= 2x.