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.

For this integral, I prefer a hyperbolic substitution $x=\tanh t$
$$\int \frac{dx}{1-x^2}=\int\frac{\operatorname{sech}^2t}{1-\tanh^2t}dt=\int dt=t+C=\operatorname{artanh}x+C=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)+C$$
You can prove the logarithmic representation for $\operatorname{artanh}t$ with this:
$$y=\operatorname{artanh}t$$
$$t=\tanh y$$
$$t=\frac{e^y-e^{-y}}{e^y+e^{-y}}=\frac{e^{2y}-1}{e^{2y}+1}$$
$$t(e^{2y}+1)=e^{2y}-1$$
Solve for $e^{2y}$, then solve for $y$
