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.
 A: By starting with $g(x) = \ln(1 + x)$ it can be determined that $g^{(n)}(x) = (-1)^{n-1} \, (n-1)! (1+x)^{-(n+1)}$ for $n \geq 1$ along with $g(0) = \ln(1 + 0) = 0$ and $g^{(n)}(0) = (-1)^{n-1} \, (n-1)!$. From this it is developed
$$\ln(1 + x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n}.$$
Now, as Mark Viola has pointed out in his comment, it can be seen that
\begin{align}
\ln(x^2 + 4) &= \ln\left( 4 \, \left(1 + \left(\frac{x}{2}\right)^{2} \right) \right) = \ln(4) + \ln\left(1 + \left(\frac{x}{2}\right)^{2} \right) \\
&= 2 \, \ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{2 n}}{4^{n} \, n} \\
&= 2 \, \ln(2) + \frac{x^{2}}{4} - \frac{x^{4}}{32} + \frac{x^{6}}{192} - \mathcal{O}(x^{8})
\end{align}
In terms of the Proposer's work: 
With
\begin{align}
\ln\left(x^2+4\right) &= \int \frac{2 \, x}{x^2+4}\,dx = \frac{1}{2} \, \int \frac{x}{\frac{x^2}{4}+1}\,dx + c_{0} \\
&= \frac{1}{2} \, \int \frac{x}{1-\left(-\frac{x^2}{4}\right)}\,dx + c_{0} \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} \, \int x \, \left(-\frac{x^2}{4}\right)^n + c_{0} \\
&= \frac{1}{2} \, \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{2n+2}}{4^{n} \, (2n+2)} + c_{0} \\
&= \frac{1}{4} \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{2n}}{4^{n-1} \, n} + c_{0} \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{2n}}{4^{n} \, n} + c_{0}.
\end{align}
The value of $c_{0}$, an essential constant of integration may be obtained by setting $x = 0$ and yields $c_{0} = \ln(4)$. This then yields the result
$$\ln(x^2 + 4) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{2n}}{4^{n} \, n} + 2 \, \ln 2. $$
A: 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
$$\ln(4+x^2) = \ln 4 + \ln\left(1+\dfrac{x^2}4\right)\approx \ln4 + \dfrac14 x^2 - \dfrac1{32}x^4+\dfrac1{192}x^6-\dots.$$
A: Your integration is not correct $$\ln\left(x^2+4\right) \ne \int \frac{1}{x^2+4}\,dx $$ You need to start with $$\ln\left(x+4\right) =  \int \frac{1}{4+x}dx$$ and go from there. 
