\arctan(x+1)'=\frac{2-2x+x^2}{4}\cdot\frac{1}{1+\frac{x^4}{4}}=\Big(\frac{1}{2}-\frac{x}{2}+\frac{x^2}{4}\Big)-\Big(\frac{x^4}{8}-\frac{x^5}{8}+\frac{x^6}{16}\Big)+\cdots,