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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

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Have you considered using integration by parts? – Shai Covo Dec 5 '10 at 8:10
What is $t$? And did you mean to define $g$ as a derivative of $f$? – Aryabhata Dec 5 '10 at 8:49
Note that the integral has a closed form in elementary functions. – Derek Jennings Dec 5 '10 at 9:40
@Shai Ok this would be easier. But it wouldn't help me study for my final :) – fdart17 Dec 5 '10 at 17:19
up vote 6 down vote accepted

Recall that $$\tan^{-1}x=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}.$$ Use this to find a power series expansion for the integrand, and integrate it termwise.

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I tried using that but then I would have needed to differentiate 3 times to get the geometric series, and then I can't integrate 3 times. Would someone be able to do this step by step? I will put up a bounty as soon as the 3 days are over. – fdart17 Dec 6 '10 at 7:30
$\tan^{-1}x^2 = \sum_{n=0}^\infty (-1)^n x^{4n+2}/(2n+1)$ so just multiply through by the $x^{11}$ and integrate. – Derek Jennings Dec 6 '10 at 8:30

$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)}$.

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I also tried sticking to this approach which didn't work out either. – fdart17 Dec 6 '10 at 7:51
This is a good approach, so for us to tell you where you went wrong, you will need to show us what you did. – GEdgar Aug 8 '11 at 13:29

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