Integration of $1/(1-x^2)$ to prove $\operatorname{arctanh}(x)$ Just for some background knowledge, I am doing this because I am trying to show that the derivative of arctanh(x) = $\frac{1}{1-x^2}$
How do I prove that the integral below is equal to arctanh(x)?
\begin{align}
\int{\frac{1}{1-x^2}dx}
\end{align}
So far, I managed to show that after integrating, it is equal to:
$$\\{\frac{1}{2}}\log\Bigl({\frac{1+x}{1-x}}\Bigl)+C$$
And since we know that, 
$$\operatorname{arctanh(x)}={\frac{1}{2}}\log\Bigl({\frac{1+x}{1-x}}\Bigl)$$
How do I show that C = 0? 
 A: If you already know that
$$
\operatorname{arctanh}(x)=\frac{1}{2}\log\frac{1+x}{1-x}
$$
then you can simply differentiate and find that
$$
\frac{d}{dx}\operatorname{arctanh}(x)=\frac{1}{1-x^2}
$$
because that's what the integral says.

You can also compute the derivative by using the definition, namely
$$
\tanh\operatorname{arctanh}(x)=x
$$
so by the chain rule
$$
1=(1-\tanh^2\operatorname{arctanh}(x))\operatorname{arctanh}'(x)
$$
and therefore
$$
\operatorname{arctanh}'(x)=\dfrac{1}{1-x^2}
$$
If you integrate the right-hand side, you find
$$
\int\dfrac{1}{1-x^2}\,dx=\frac{1}{2}\log\frac{1+x}{1-x}+c
$$
and therefore $\operatorname{arctanh}(x)$ and this function differ by a constant
$$
\operatorname{arctanh}(x)=\frac{1}{2}\log\frac{1+x}{1-x}+c
$$
Evaluating at $0$ shows the constant is $0$.
A: Note that  you can directly show  $\operatorname{argtanh}x=\frac12\ln\frac{1-x}{1+x}$, by solving the equation
$$x=\tanh y=\frac{\mathrm e^{2y}-1}{\mathrm e^{2y}+1}\iff x(\mathrm e^{2y}+1)=\mathrm e^{2y}-1\iff \mathrm e^{2y}=\frac{1+x}{1-x}.$$
Also, you can easily get the derivative with the generic formula for the derivative of an inverse function:
$$\operatorname{argtanh}'(x)=\frac1{\tanh'(\operatorname{argtanh}x)}=\frac 1{1-\tanh^2(\operatorname{argtanh}x)}=\frac1{1-x^2}$$
A: Set $x=0$ and since $\operatorname{arctanh} x=0$, then$$\begin{align*}\operatorname{arctanh}0 & =\frac 12\log\left(\frac {1+0}{1-0}\right)+C\\0 & =0 + C\end{align*}$$Thus $C=0$
