Integration by parts for $I_n=\int \frac 1 {(x^2-1) ^n} d x$ and relation with $I_1$ Compute $I_n=\int \frac 1 {(x^2-1) ^n} \Bbb d x$.
My work: $I_1 = \int \frac 1 {x^2-1} \Bbb d x= \frac 1 2 \int (\frac 1 {x-1} -\frac 1 {x+1}) \Bbb d x = \frac 1 2 \ln (x-1) - \frac 1 2 \ln (x+1)$
 A: I suspect that you are looking for a solution at an elementary level; such a thing does not exist and your integral does not have a simple solution in closed form. The result involves the hypergeometric function that you most probably have not studied.
For your curiosity, Mathematica 7 produces the following answer: $x \Big( \frac {1-x^2} {-1 + x^2} \Big) ^n {}_2 F _1 (\frac 1 2, n, \frac 3 2, x^2)$, where ${}_2 F _1$ is the hypergeometric function.
A: There is a relevant Comment from a Question closed as a dupe of this one, which I promote to an answer here. By integration by parts, one obtains for $m\ge 0$,
\begin{align}\int\frac{dt}{(1-t^2)^m}&=\frac{t}{(1-t^2)^m}-2m\int\frac{t^2}{(1-t^2)^{m+1}}dt\\&=\frac{t}{(1-t^2)^m}+2m\int\frac{dt}{(1-t^2)^m}-2m\int\frac{dt}{(1-t^2)^{m+1}},\end{align}
i.e.
$$ I_m = \frac{t}{(1-t^2)^m} + 2m I_m - 2m I_{m+1}.$$
So there is a reduction formula
$$ I_{m+1} = \frac{2m-1}{2m}I_m + \frac t{2m(1-t^2)^m}$$
which can be used inductively  to write out a formula for $I_m$.
