Proving that $\int \frac{dx}{(1-x^2)}$ equals the inverse hyperbolic tangent of $x$ I prove the following fact
$$\int\limits \dfrac{dx}{(1-x^2)}=\tanh^{-1}x$$
I show, by integrating by substitution, that the integral equals 
$$- \dfrac{\ln(1-x^2)}{2x}.$$
Setting $x=\tanh z$, we get $$-\dfrac{\ln\operatorname{sech} z}{\tanh z}.$$
Then, since $z=\tanh^{-1}x$, the above equation yields
$$-\dfrac{\ln(1-x^2)}{2x}.$$
Is this sufficient to prove the first equation?
 A: Your integration is incorrect; the derivative of $-\frac{\ln(1-x^{2})}{2x}$ is $$-\frac{\frac{2x}{1-x^{2}}-2\ln(1-x^{2})}{4x^{2}}$$
The correct way to proceed is by partial fractions; notice the following:
$$\frac{1}{1-x^{2}}=\frac{1}{(1-x)(1+x)}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$$
And so, integrating this, we get$$\int\frac{1}{1-x^{2}}dx=\int\frac{1}{2(1-x)}+\int\frac{1}{2(1+x)}=\frac{1}{2}\left(-\ln(1-x)+\ln(1+x)\right)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$$
Which is another expression for $\tanh^{-1}(x)$ - are you familiar with this?   
As an aside, I think I can see where your incorrect expression has come from, and it is important to see that it is wrong. When integrating by substitution, you appear to know that we must compute the derivative of our substitution - in this case, the derivative $1-x^{2}$ is $-2x$. However, we cannot bring this outside of the integral.
A: By integrating by method of partial fractions, you will see that the result is not correct. I think you can use the definition of $\tanh x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=1-\frac{2e^{-x}}{e^{x}+e^{-x}}=y$ instead. Do the right integration first, and then find $y$ with respect to $x$. Note that $|x|<1$.
