What is $\int \frac{-2}{x^2-1}\,dx$

Question

For $x\in\mathbb{R}$, what is the value of $$ \int \frac{-2}{x^2-1}\,dx? $$ Using partial fractions, we get $$ \int \frac{-2}{x^2-1}\,dx= \int \frac{1}{x+1}\,dx+ \int \frac{-1}{x-1}\,dx=\log|x+1|-\log|1-x|+C=\log \left|\frac{x+1}{1-x} \right|+C $$ for all $x\in\mathbb{R}\backslash \{-1,1\}$. However, using the substitution $x = \tanh u$, we get $$ \int \frac{-2}{x^2-1}\,dx= -2\int\frac{\text{sech}^2 u}{\tanh^2u-1} \,du=2u+C=2\tanh^{-1} x+C $$ which only hold for $|x|<1$. For this interval, the identity $2\tanh^{-1} x=\log\left( \frac{1+x}{1-x} \right)$ holds. However, I believe that only the first answer is correct, for all $x\in\mathbb{R}\backslash \{-1,1\}$. What am I missing here? Is it because this substitution only holds for $|x|<1$ and therefore I am restricting the evaluation of the integral to this domain? Or are these answers equivalent up to a constant?

Answer 1

Since $\exp (x \pm \pi i/2) = \pm ie^x$, we have \begin{align*} \tanh \left(x + \frac{\pi i}{2}\right) = \frac{ie^{x} + i e^{-x}}{i e^{-x} - i e^{-x}} = \frac{1}{\tanh x}; \end{align*} that is, \begin{align*} \tanh^{-1} \frac{1}{x} = \frac{\pi i}{2} + \tanh^{-1} x. \end{align*} (I'm glossing over the matter of defining the range of $\tanh^{-1}$ properly, since $\tanh$ has period $\pi i$.) We also have \begin{align*} \left|\frac{1 + x^{-1}}{1 - x^{-1}}\right| = \left|\frac{x+1}{x - 1}\right| = \left|\frac{1+x}{1-x}\right| \end{align*} for $x\not = 0, 1$. The upshot is that we can reconcile the two integrands by taking $x \to 1/x$ and, up to a constant of integration, extend the result for $|x| < 1$ to all real $x\not = -1, 1$. This constant of integration is going to be complex, but that's because the substitution $x \to \tanh^{-1} x$ is only real-valued for $|x| < 1$.