How can I develop $f(x) = (x^2+1)\cdot \arctan(x)$ into a Taylor series about $x_{0} = 0$? How can I develop $f(x) = (x^2+1)\cdot \arctan(x)$ into a Taylor series about $x_{0} = 0$ ?
I see that $f(x) = \frac{\arctan(x)}{\arctan'(x)}$ and I also know what the Taylor expansion of $\arctan(x)$ looks like about zero. But I don't know how to develope what they wanted. I get a series that is not a power series... 
 A: The key here is that you want the same power of $x$ for all the terms of the sum.
\begin{align*}
(x^2+1)\arctan(x) & = (x^2+1) \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\
& =  \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}+\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\
& = \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}+\sum_{k=1}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}+x\\
& =\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}-\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+3}+x\\
& =x+\sum_{k=0}^{+\infty}(-1)^k \left(\frac{1}{2k+1}-\frac{1}{2k+3}\right)x^{2k+3} 
\end{align*}
A: Simply multiply each term in $\arctan(x)$ by $x^2+1$, then collect like terms.
A: $$f'(x)=2x\tan^{-1}x+1$$
$$f'(x)=1+2x\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n+1}}}{(2n-1)}}\,{{x}^{2n-1}}=1+2\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n+1}}}{(2n-1)}}\,{{x}^{2n}}$$ therefore
$$f(x)=c+x+2\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n+1}}}{(2n+1)(2n-1)}}\,{{x}^{2n+1}}$$ 
we have $f(0)=0$, thus $c=0$ and we have
$$f(x)=x+2\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n+1}}}{(2n+1)(2n-1)}}\,{{x}^{2n+1}}$$
