f(x) = [integral from 0 to x] tan−1(t)dt based at b = 0.
$\displaystyle \frac{1}{1+t^2} = \frac{1}{1 - (-t^2)} = 1 - t^2 + t^4 - t^6 + ...$
integrate term for term ...
$\displaystyle \arctan{t} = C + t - \frac{t^3}{3} + \frac{t^5}{5} - \frac{t^7}{7} + ...$
since $\displaystyle \arctan(0) = 0$ ... $\displaystyle C = 0$
$\displaystyle \arctan{t} = t - \frac{t^3}{3} + \frac{t^5}{5} - \frac{t^7}{7} + ...$
o.k. ... integrate $\displaystyle \arctan{t}$