# Find the taylor series about x=0 for the following function g(x) = ln(x^(2) +1)

*print*Print*list*Cite

If we try to take derivative after derivative, things will get messy quickly. Instead, we can use the following trick:

`g(x)= "ln" (x^2+1)`

`g'(x)= (2x)/(x^2+1)`

We can find the power series for `g'(x)` using what we know about geometric series:

`1+z+z^2+z^3+... = (1)/(1-z)` (provided that `|z|<1` )

If we say `z=-x^2`, we have:

`1-x^2+x^4-x^6+...=(1)/(1+x^2)`

So:

`g'(x)= (2x)/(x^2+1)`

`=2x(1-x^2+x^4-x^6+...)`

`=2x-2x^3+2x^5-2x^7+...`

We can integrate term by term to get `g(x)`

`g(x) = C + x^2 - (2/4)x^4 + (2/6)x^6-(2/8)x^8+...`

`=C+x^2 - (x^4)/2 + (x^6)/3 - (x^8)/4 + ...`

To get C, we plug in `x=0`:

`g(0) = C + 0^2 + ...`

`g(0) = "ln" (0^2+1) = 0`

So `C=0`

And

`g(x)=x^2 - (x^4)/2 + (x^6)/3 - (x^8)/4 + ...`

Or, in summation notation:

`g(x) = sum_(n=1)^oo (x^(2n))/n`