Power series of $f(x)=\ln (x^2+4)$ I am supposed to find a power series representation of
$$f(x)=\ln\left(x^{2}+4\right).$$
Then, I am to graph it and observe what happens as $n$ increases. My attempt at a solution:
$$\ln\left(x^2+4\right) = \int \frac{1}{x^2+4}\,dx = \frac{1}{4}\int \frac{1}{\frac{x^2}{4}+1}\,dx = \frac{1}{4}\int \frac{1}{1-\left(-\frac{x^2}{4}\right)}\,dx.$$
Now that it is in the $\frac{1}{1-x}$ format, the power series representation is $\sum_{n=0}^{\infty} \int \frac{1}{4}\left(-\frac{x^2}{4}\right)^n$. The first terms are
$$\frac{1}{4} \left(x-\frac{x^5}{80}+\frac{x^7}{448}-\frac{x^9}{2304}+\cdots\right).$$
But as I graph these, they look nothing like the graph of $f(x)=\ln(x^2 +4)$. I am not sure if I turned the revised formula into a sum correctly.

At first: the cubic member missed in OP, and the signs of the other ones inverted. So the series is wrong at all.
Corrrect realizaion of  the OP idea is:
$$\ln(4+x^2) = \dfrac14x - \dfrac1{48}x^3+ \dfrac1{320}x^5- \dfrac1{1792}x^7+\dots.$$
Then, there are known the series
$$\ln(1+t) = t-\dfrac12t^2+\dfrac13t^3+\dots,\tag1$$
which converges only if $|t|<1.$
Then
$$\ln(4+x^2) = \ln 4 + \ln\left(1+\dfrac{x^2}4\right)\approx \ln4 + \dfrac14 x^2 - \dfrac1{32}x^4+\dfrac1{192}x^6-\dots.$$
