Find Maclaurin series and 8-th derivative at $0$ for $ln(4+x^2)$

Question

So I evaluated the Maclaurin series for that and end up with $\sum (-1)^{n-1}\frac{(x^2+3)^{n}}{n}$ and my problem is I have no idea how to get $f^{(8)}(0)$ because of that 3 constant term.

Answer 1

You have not found the Maclaurin series of $\log(4+x^2)$, you have found a power series in $x^2+3$.
The former is given by $$ f(x)=2\log(2)+\log\left(1+\frac{x^2}{4}\right) = 2\log 2+\sum_{n\geq 1}\frac{(-1)^{n+1}}{n 4^n} x^{2n} $$ from which it follows that $$ f^{(8)}(0) = 8!\cdot\frac{(-1)^{4+1}}{4\cdot 4^4} = -\frac{315}{8}.$$