Calculus Integral from Partial Fractions When you have an irreducible quadratic factor repeated you can get integrals that look like $\int \dfrac{dx}{(x^2+a)^m}$, where $m>1$, integer, and $a>0$.  What is the best way to integrate this function?  Is there more than one way?

The recurrence relation derived by Jack D'Aurizio in his answer above can actually be solved explicitly for $I_m$. First let me quote it below, but changing the recurrence index to $n$:

$$
I_{n+1} = \frac{2n+1}{2n}\,I_{n} + f_n(t)
\qquad (n \ge 1) \qquad (1)
$$
  where
  $$
\begin{align}
I_0 &= t \\[0.05in]
I_1 &= \arctan(t) \\[0.05in]
f_n(t) &\equiv \frac{1}{2n}\frac{t}{(1+t^2)^n} \qquad (n \ge 1)
\end{align}
$$

Now define the quantity

$$
P_{n} \equiv \prod_{k\,=\,1}^{n} \left(\frac{2k+1}{2k}\right)
           = \prod_{k\,=\,1}^{n} \left(1 + \frac{1}{2k}\right)
$$

Then, dividing (1) by $P_n$, we find
$$
\frac{I_{n+1}}{P_{n}} - \left(\frac{2n+1}{2n}\right)\frac{I_{n}}{P_{n}} = \frac{f_n}{P_{n}}
$$
Since
$$
P_{n} = \left(\frac{2n+1}{2n}\right)P_{n-1}
$$
we have
$$
\frac{I_{n+1}}{P_{n}} - \frac{I_{n}}{P_{n-1}} = \frac{f_n}{P_{n}}
$$
Now sum both sides on $n$:
$$
\sum_{n\,=\,2}^{m-1} \left( \frac{I_{n+1}}{P_{n}} - \frac{I_{n}}{P_{n-1}} \right)= \sum_{n\,=\,2}^{m-1} \frac{f_n}{P_{n}}
$$
The LHS telescopes, so
$$
\frac{I_{m}}{P_{m-1}} - \frac{I_{2}}{P_{1}} = \sum_{n\,=\,2}^{m-1} \frac{f_n}{P_{n}}
$$
and
$$
I_{m} = P_{m-1} \left(
\frac{I_{2}}{P_{1}} + \sum_{n\,=\,2}^{m-1} \frac{f_n}{P_{n}} \right) = P_{m-1} \left(
\frac{I_{2}}{P_{1}} - \frac{f_1}{P_{1}} + \sum_{n\,=\,1}^{m-1} \frac{f_n}{P_{n}} \right) = P_{m-1} \left(
I_1 + \sum_{n\,=\,1}^{m-1} \frac{f_n}{P_{n}} \right)
$$
where (1) was used in the last equality to remove $I_2$.
Hence,

$$
I_{m} = P_{m-1} \left(
\arctan(t) + \sum_{n\,=\,1}^{m-1} \frac{t\,(1+t^2)^{-n}}{2n\,P_{n}} \right)
\qquad (m \ge 2)
$$

