Find the recursive integral of $I\left(n\right)=\int \frac{1}{\left(1-x^2\right)^n}\:dx$ Using only simple calculus and rewriting the integral how could I find the recursive integral of
$I\left(n\right)=\int \frac{1}{\left(1-x^2\right)^n}\:dx$
I tried using trig substitution like so:
$x = \sin(u)$ with $dx = \cos(u) du$
writing $1 = \sin^2(u) + \cos^2(u)$
and also $(1-\sin(u)^2)=\cos^2(u)$
Eventually I reach that $I(n) = \int \frac{\sin^2\left(u\right)}{\cos^{n-1}du}du + \int \cos^{1-n}\left(u\right)du$
I stopped here as I do not know how to continue or if the route I chose is even ok.

For $n \ge 2$ we have
$$\begin{align}\int\frac{1}{(1-x^2)^n}dx&=\int\frac{1-x^2+x^2}{(1-x^2)^n}dx\\
&=\int\frac{1}{(1-x^2)^{n-1}}dx+\int x\cdot\frac{x}{(1-x^2)^n}dx\end{align}.$$
For the second integral, use integration by parts with $u=x, dv=\frac{x}{(1-x^2)^n}dx$:
$$\int x\cdot\frac{x}{(1-x^2)^n}dx=x\cdot\frac{1}{2(n-1)(1-x^2)^{n-1}}-\frac{1}{2(n-1)}\int\frac{1}{(1-x^2)^{n-1}}dx $$
so we get
$$\begin{align}I(n)&=I(n-1)+\frac{x}{2(n-1)(1-x^2)^{n-1}}-\frac{1}{2(n-1)}I(n-1)\\
&=\frac{2n-3}{2(n-1)}I(n-1)+\frac{x}{2(n-1)(1-x^2)^{n-1}}\end{align} $$
