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In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.

From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}{x-a}) $ [as we know $\ln(\frac{1}{x})=-\ln(x)$]

But, in my book the integration of $\frac{1}{a^2-x^2}$ is written as $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $.

We know that the integration in the second case cannot be $\frac{1}{2a}\ln(\frac{x+a}{x-a}) $ and $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $ at the same time. Which formula is the correct one and why?

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    there is an absolute sign in the formula : http://www.petervis.com/mathematics/integration_formula_sheet/integration_formula_sheet/integration_formula_sheet_033.gif – 2017-01-04
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    Why should there be an absolute sign ? @shrey – 2017-01-04
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    because it is here also : http://mathematicsi.com/wp-content/uploads/2011/06/lnx-integration-rule2.png – 2017-01-04
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    Yes, they are both correct. Or, rather, the one that is defined is correct. Remember that you can only take logarithms of positive numbers. Because $(x+a)/(x-a)$ and $(x+a)/(a-x)$ are negatives of each other at most one of them is positive. Conclusion: which formula to use depends on the range of $x$, and where it lies in relation to $\pm a$. – 2017-01-04
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    @JyrkiLahtonen That does make sense. Thanks :) I got it now! – 2017-01-04

4 Answers 4

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Let $f(x)=\frac{1}{x^2-a^2}$. $f$ is continuous at $(-\infty,-|a|),(-|a|,|a|)$ and $(|a|,+\infty)$.

in each interval $$2af(x)=\frac{1}{x-a}-\frac{1}{x+a}$$

and $$\int f(x)dx=\frac{1}{2a}\ln(\frac{|x-a|}{|x+a|}).$$

the final expression depends on which interval $J$ we want the antiderivative. for example, if $a>0$ and $J=(-a,a)$, we have

$$\int f(x)dx=\frac{1}{2a}\ln( \frac{a-x}{x+a} )+C$$

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    This answer does not address the case of $1/(a^2-x^2)$. – 2017-01-04
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    This is a suspicious downvoting. Isn't it? – 2017-01-04
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    It's mine. I noticed that you don't address the main question in the OP. – 2017-01-04
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    @YvesDaoust The OP must speak about the intervals where the antiderivative exist. – 2017-01-04
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    @YvesDaoust Didn't you see the first line. – 2017-01-04
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We do have that $$I =\int \frac {1}{a^2-x^2} dx =\frac {1}{2a}\ln \frac {x+a}{x-a} $$ But however we apply the absolute value function to arguments of logarithm functions in order to extend the antiderivative's domain giving us: $$I =\int \frac {1}{a^2-x^2} dx =\frac {1}{2a} \ln |\frac {x+a}{x-a}|$$Hope it helps.

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    What does "extend the antiderivative's domain" mean ? Can you give an example ? – 2017-01-04
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    @Hatheat Take $x \in (-a,a) $. For these $x $ the logarithm is negative. – 2017-01-04
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Neither is correct.

It's not restrictive to assume $a>0$ (for $a=0$ the antiderivative has a different form; for $a<0$ change the intervals below accordingly).

The function $\frac{1}{x^2-a^2}$ is defined for $x\ne\pm a$ and is negative over $(-a,a)$, positive over $(-\infty,-a)$ and $(a,\infty)$.

The sign of $\frac{x-a}{x+a}$ is the same as the sign of $x^2-a^2$ (for $x\ne\pm a$, of course), so it's clear that $$ \frac{1}{2a}\ln\frac{x-a}{x+a} $$ cannot be an antiderivative over $(-a,a)$, because it is undefined there.

There's a simple way out. An antiderivative of $1/x$ is $\ln|x|$ (up to an arbitrary constant over $(-\infty,0)$ and an arbitrary constant on $(0,\infty)$, with no connection between each other).

The partial fraction decomposition is $$ \frac{1}{x^2-a^2}=\frac{1}{2a}\frac{1}{x-a}-\frac{1}{2a}\frac{1}{x+a} $$ so we can write $$ \int\frac{1}{x^2-a^2}\,dx= \frac{1}{2a}(\ln|x-a|-\ln|x+a|)= \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| $$ up to arbitrary constants in the intervals $(-\infty,-a)$, $(-a,a)$ and $(a,\infty)$.

Since $1/(a^2-x^2)=-1/(x^2-a^2)$, we have $$ \int\frac{1}{a^2-x^2}\,dx= -\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|= \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|^{-1}= \frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right| $$ again up to constants as before.

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    Suppose $a$ is given to be negative in the question, however in the integral when $a^2$ is written it becomes positive. But, when we write the $1/2a$ in the final answer, will we put the actual negative value of $a$ or will be put the absolute value of $a$ ? (Nice answer btw :)) – 2017-01-04
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    @Hatheat I was supposing $a>0$, but it's better to add it upfront. Note that only the intervals change form, not the integral itself. – 2017-01-04
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WLOG, $a>0$. As there are singularities at $x=-a$ and $x=a$, the expressions can differ in the intervals $(-\infty,-a)$, $(-a,a)$ and $(a,\infty)$, depending on the signs of $x+a$ and $x-a$.

Correct expressions are

$$\int\frac{dx}{x^2-a^2}=\frac1{2a}\log\left|\frac{x+a}{x-a}\right|=\frac1{2a}\log\left|\frac{a+x}{a-x}\right|.$$

(But you are not allowed to evaluate the definite integrals across the singularities.)

Then the effect of a change of sign is

$$\int\frac{dx}{a^2-x^2}=-\frac1{2a}\log\left|\frac{x+a}{x-a}\right|=\frac1{2a}\log\left|\frac{x-a}{x+a}\right|=\frac1{2a}\log\left|\frac{a-x}{a+x}\right|.$$

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    This is the best answer till now. Thanks for including the "definite integral" comment ! – 2017-01-04
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    @SalahFatima: fixed. – 2017-01-04
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    I am facing a new confusion now. Suppose $a$ is given to be negative in the question, however in the integral when $a^2$ is written it becomes positive. But, when we write the $\frac{1}{2a}$ in the final answer, will we put the actual negative value of $a$ or will be put the absolute value of $a$ ? @Yves – 2017-01-04