Integrate $\int\frac{dx}{a^2-x^2}$

Question

For what values of $x$ is this valid?

$$\int\frac{dx}{a^2-x^2}=\frac{1}{a}\tanh^{-1}\frac{x}{a}+C$$

I think the anwer should be $-a<x<a$ because of the domain of $tanh^{-1}$. Is this correct?

Thanks!

Answer 1

It makes sense to divide the problem into finding an antiderivative in the three intervals $I_{-}=(-\infty,-a)$, $I_0=(-a,a)$ and $I_+=(a,\infty)$ (I assume, which is not restrictive, that $a>0$).

With a simple substitution, the problem for $I_{-}$ can be reduced to the problem for $I_+$.

If we are in $I_0$, we can make the substitution $x=a\tanh t$, so the integral becomes $$ \int \frac{1}{a^2(1-\tanh^2t)}a(1-\tanh^2t)\,dt=\frac{t}{a}+C_0 $$ and so an antiderivative is $$ \frac{1}{a}\operatorname{artanh}\frac{x}{a}+C_0 $$

If we are in $I_+$ the substitution above cannot be done, but we can do instead $x=a/\tanh t$ and we get $$ \int\frac{1}{a^2\left(1-\dfrac{1}{\tanh^2t}\right)} \left(-\frac{a}{\tanh^2t}(1-\tanh^2t)\right)dt =\frac{t}{a}+C_+ $$ so the antiderivative is $$ \frac{1}{a}\operatorname{artanh}\frac{a}{x}+C_+ $$

In summary, the most general antiderivative is $$ \begin{cases} \dfrac{1}{a}\operatorname{artanh}\dfrac{a}{x}+C_- & x<-a \\[6px] \dfrac{1}{a}\operatorname{artanh}\dfrac{x}{a}+C_0 & -a<x<a \\[6px] \dfrac{1}{a}\operatorname{artanh}\dfrac{a}{x}+C_+ & x>a \end{cases} $$

Of course, rewriting the integral as $$ \frac{1}{2a}\int\left(\frac{1}{a-x}+\frac{1}{a+x}\right)dx $$ is easier.