Integration of $1/(1-x^2)$ to prove $\operatorname{arctanh}(x)$

Question

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?

Answer 1

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.

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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 $0$.

Answer 2

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: $$\operatorname{argtanh}'(x)=\frac1{\tanh'(\operatorname{argtanh}x)}=\frac 1{1-\tanh^2(\operatorname{argtanh}x)}=\frac1{1-x^2}$$

Answer 3

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 $C=0$