Help Getting Started on Evaluating this Integral as a Power Series

Question

I'm studying for the Math Subject GRE this fall, so I'm reviewing everything in Stewart's Calculus - Early Transcendentals 6th Ed. I've hit a brick wall with this problem, from the first section about power series representations of functions, in which I'm supposed to evaluate the following as a power series:

$$\int \dfrac{x - \arctan(x)}{x^3}dx$$

I don't know how to evaluate the $\dfrac{1}{x^2}$ term as a power series, nor do I know what to do with $\dfrac{\arctan (x)}{x^3}$, since the lowest term in the series expansion for arctangent is $x$.

I suspect there must be some trick to this that I'm not seeing, so any help getting me started would be greatly appreciated!

First, find the power series for $x-\arctan(x)$ then divide by $x^3$. — John Wayland Bales, Aug 02 '19 at 18:21
Oh...man. Well, I'd rather feel dumb now rather than later. Thank you! — spectacularlyrubbish, Aug 02 '19 at 18:24

score 1 · Answer 1 · answered Aug 02 '19 at 18:27

\begin{align} \int \dfrac{x - \arctan(x)}{x^3}dx &= \int \dfrac{x - \sum_{k=0}^\infty{\frac{(-1)^kx^{2k+1}}{2k+1}}}{x^3}dx\\ &= \int \dfrac{x - (x + \sum_{k=1}^\infty{\frac{(-1)^kx^{2k+1}}{2k+1}})}{x^3}dx\\ &= -\int \dfrac{\sum_{k=1}^\infty{\frac{(-1)^kx^{2k+1}}{2k+1}}}{x^3}dx\\ &= -\int \sum_{k=1}^\infty{\frac{(-1)^kx^{2k-2}}{2k+1}} dx\\ &= -\sum_{k=1}^\infty \frac{(-1)^k}{2k+1}\int x^{2k-2} dx \end{align}

Note: The step where I swapped the integral and summation operators requires some condition to be true.

Help Getting Started on Evaluating this Integral as a Power Series

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