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Question:

Find the antiderivative(s): $\int{\frac{1}{x^2-a^2}}\mathbb{d}x$

My book's attempt:

$$\int{\frac{1}{x^2-a^2}}\mathbb{d}x$$

$$\int{\frac{1}{(x+a)(x-a)}}\mathbb{d}x$$

$$\frac{1}{2a}\int\left({\frac{1}{x-a}-\frac{1}{x+a}}\right)\mathbb{d}x$$

$$\frac{1}{2a}[\ln(x-a)-\ln(x+a)]+C$$

$$\frac{1}{2a}\ln(|\frac{x-a}{x+a}|)+C,x\ne\pm a$$

$\frac{1}{x^2-a^2}$ is a discontinuous function, and there are three integration domains of it: $(-\infty, -a)$, $(-a,a)$ & $(a,+\infty)$. We will get three different family of functions as antiderivatives for the three different integration domains of this function, and two antiderivatives belonging to two different integration domains/family of functions will not differ by a constant.

My question is that my book found out the antiderivative belonging to which integration domain?

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tryingtobeastoic
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    The antiderivative is generic and works in any of those three intervals you need to work with. In any of those open intervals, if you differntiate the latter function, you'll get the initial rational function – Luciano Jan 27 '22 at 17:05
  • @Luciano That is not how that works, though. The antiderivative of a function must be specified on the entire domain of the function, not an open connected subset thereof. So the three constants of antidifferentiation must be given. The correct answer should always be given as $$\begin{cases}\ln\left(\frac{x-a}{x+a}\right)+C_0&x\lt-a\\ln\left(-\frac{x-a}{x+a}\right)+C_1&-a\lt{x}\lt{a}\\ln\left(\frac{x-a}{x+a}\right)+C_2&a\lt{x}\end{cases}.$$ Giving it as the book did, without specifying a restriction of the domain, is simply incorrect. – Angel Jan 27 '22 at 17:17
  • @Angel

    $$ \begin{cases} \frac{1}{2a}\ln(\frac{x-a}{x+a})+C_0,x<-a\ \frac{1}{2a}\ln(-\frac{x-a}{x+a})+C_1,-a<x<a\ \frac{1}{2a}\ln(\frac{x-a}{x+a})+C_2,a<x \end{cases} $$

    Shouldn't you have written this?

     – tryingtobeastoic Jan 28 '22 at 05:33
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    @tryingtobeastoic Yes, that is what I meant. – Angel Jan 28 '22 at 12:53

2 Answers2

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Your book used the absolute value in logarithm and this makes the antiderivative having a unique expression. It gives the same formula for the three domains.

If your integrand was $$f(x)=\frac{1}{\sqrt{|x^2-1|}}$$ You won't get the same expression. in each of the intervalls $(-\infty,-1) \; (-1,1)\;, (1,\infty) $,

there is a special antiderivative.

hamam_Abdallah
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  • Yes, the answer given does differentiate to the correct answer, but the problem is that the answer given is not the complete family of antiderivatives, so it is not actually correct as an answer. – Angel Jan 27 '22 at 17:27
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If the component pieces of a function are governed by the same 'rule' that has a primitive (e.g., this is not so in the case of $f(x)=|\frac1x|),$ then, from the Fundamental Theorem of Calculus, we do expect their primitives to differ by a constant.

(It is accurate to say that on each integration interval, $\frac1x$ has primitive $\ln|x|.)$

In the given solution, there has been no additional condition imposed on the integrand (to apply a certain law/theorem, for example) on top of the starting restriction $x\ne\pm a,$ so there's no basis for inferring that the result doesn't apply on $(-\infty,-a)$ or $(-a,a)$ or $(a,\infty).$

ryang
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