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.
 A: Partial fraction here is the best way, because
$$\frac{1}{1-x^2}=\frac{-1/2}{1-x}+\frac{1/2}{1+x}.$$
Then we have linear factor and it is not difficult to evaluate.
If you want a different method, you can consider write
$$\frac{1}{1-x^2}=\frac{1}{(\sqrt{1-x^2})^2}$$
and then the trigonometric substitution $u(x)=\sin x$.
The way using integration by parts as you noticed seems really impractical, if you work with the second antiderivative you eventually you might be interested in $\int \frac{x^4}{(1-x^2)^3}\, dx$ and you don't get out of that growth of powers. At some point we might decide to stop you and use another method, but that wouldn't make much sense.
I guess they tried to say "partial fractions" in the video as the question suggested by the professor Hans in the comments.
A: One alternative is to render $u=\frac{1+x}{1-x}$, matching the zero and pole of $u$ with poles of the integrand.
Then $x=\frac{u-1}{u+1}, dx=\frac{2du}{(u+1)^2}$
and so
$$\dfrac{dx}{1-x^2}=\dfrac{2\left(\dfrac{du}{(u+1)^2}\right)}{1-\left(\dfrac{(u-1)^2}{(u+1)^2}\right)}=\dfrac{2du}{(u+1)^2-(u-1)^2}=\dfrac{du}{2u}$$
from which the antiderivative may he read off via the definition of the natural logarithm as
$$\dfrac12\ln u+C=\dfrac12\ln\left(\dfrac{1+x}{1-x}\right)+C$$
A: 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}
A: 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 $y$
A: Trigonometric substitution also works
$$
\begin{aligned}
\int \frac{d x}{1-x^2}& \stackrel{x=sin\theta}{=} \int \frac{\cos \theta d \theta}{\cos ^2 \theta} \\
& =\int \sec \theta d \theta \\
& =\ln |\sec \theta+\tan \theta|+C \\
& =\ln \left|\frac{1+x}{\sqrt{1-x^2}}\right|+C \\
& =\ln \left|\sqrt{\frac{1+x}{1-x}}\right| \\
& =\frac{1}{2} \ln \left|\frac{1+x}{1-x}\right|+C
\end{aligned}
$$
