Evaluate the integral as a power series:
$\displaystyle\int x^{11}\cdot\tan^{-1}(x^2)~\mathrm dx$
We have been using Abel's theorem to do this (and the fact that the function is differentiable and integrable on it's interval of convergence in this case) From what I can tell, an approach to this question would be to find a closed form similar to $\dfrac1{1-x}$ by taking the derivative of $\tan^{-1}(x)$.
so let $f(x) = \int x^{11}\cdot\tan^{-1}(x^2)~\mathrm dx$
then let $t = x^2, g(t) = f^{~\prime}(x^2)$
then divide by $x^{11/2}$,
$\dfrac{g(t)}{t^{11/2}} = \tan^{-1}(t)$
then take the derivative of both sides so
derivative of $\displaystyle\frac{g(t)}{t^{11/2}} = \frac1{t^2+1}$
But I am stuck here relating this to $\dfrac1{1-x}$. Maybe I am looking at this the wrong way. Any help would be appreciated. Thanks