Can $\frac{1}{1-x^2}$ be integrated by parts? I was watching a video of a professor who said that $\int \frac{1}{1-x^2} dx$ can be done using the method of integration by parts. Though he skipped the solution and simply gave the answer, I tried solving it myself but I'm having difficulty moving forward. What I did was:
Let $u = \frac{1}{1-x^2}, du = \frac{2x}{(1-x^2)^2} dx , dv = dx, v=x$.
Then,
$$\int \frac{dx}{1-x^2} = \frac{x}{1-x^2} - \int \frac{2x^2}{(1-x^2)^2} dx$$
However, the new integral seems problematic. I tried doing the method again but the equation just ends up to be $0=0$.
Is integration by parts not a viable way of solving this integral?
P.S.: I already know the answer using a different method (partial fraction decomposition). I just want to know how to solve the integral in a different way.

Here is to integrate by parts, albeit unnaturally
\begin{align}
I=\int \frac{dx}{1-x^2}
=\int\ln^{\frac12}\frac{1+x}{1-x}d\left(\ln^{\frac12}\frac{1+x}{1-x}\right)
=\ln\frac{1+x}{1-x}-I=\frac12 \ln\frac{1+x}{1-x}
\end{align}
