10th derivative of a function

\begin{align} f(x) &= \ln(2 + x^{2}) = \ln 2 + \ln\left( 1 + \frac{x^{2}}{2}\right) \\ &= \ln 2 + \sum_{k=1}^{\infty} \frac{(-1)^{k-1} \, x^{2k}}{2^{k} \, k} \\ &= \ln 2 + \frac{x^{2}}{2} - \frac{x^{4}}{8} + \frac{x^{6}}{24} - \frac{x^{8}}{64} + \frac{x^{10}}{160} - \cdots \\ f(x) &= \ln 2 + \frac{x^{2}}{2!} - 3 \, \frac{x^{4}}{4!} + 30 \, \frac{x^{6}}{6!} - 630 \, \frac{x^{8}}{8!} + 22680 \, \frac{x^{10}}{(10)!} - 1247400 \, \frac{x^{12}}{(12)!} + \cdots \end{align} From this series expansion it can be determined that \begin{align} f^{(2n+1)}(0) &= 0 \hspace{10mm} \text{for} \quad n \geq 0 \\ f^{(2n)}(0) &= \frac{(-1)^{n-1} \, (2n)!}{2^{n} \, n} = \frac{(-1)^{n-1} \, (2n-1)!}{2^{n-1}} \hspace{10mm} \text{for} \quad n \geq 1 \\ f(0) &= \ln 2. \end{align}

We have $$ \frac{2x}{x^2+2} = \frac{x}{1+\frac{x^2}{2}} = x\bigg(1-\frac{x^2}{2} + \frac{x^4}{4} - \frac{x^6}{8} + \frac{x^8}{16}+\cdots \bigg). $$ Then integrating we get $$ \log(x^2+2)-\log2 = \int_0^x \frac{2t}{t^2+2}\; dt = \frac{x^2}{2}-\frac{x^4}{2\cdot 4} + \frac{x^6}{4\cdot 6} - \frac{x^8}{8\cdot 8} + \frac{x^{10}}{10\cdot 16}+\ldots $$

Answer: $22680$ :)

\begin{align} f'(x)=\frac{x}{1+\frac{x^{2}}{2}}=\sum_{k=0}^{\infty }(-1)^{k}\left ( \frac{x^{2k+1}}{2^{k}} \right )=x-\frac{x^{3}}{2}+\cdots+\frac{x^{9}}{2^{4}}-\frac{x^{11}}{2^{5}}+\cdots \end{align}

Therefore if we differentiate this series nine times, we see that $f^{(10)}(0)$ is the constant term, which is $\frac{9!}{16}=22680$